NCERT Solutions · Class 12 Chemistry Chemical Kinetics

For a gas-phase reaction run at constant volume and temperature, pressure is directly proportional to moles — so you can track the reaction using partial pressures exactly as you would use concentrations.The reaction is \[SO_2Cl_2(g) \rightarrow SO_2(g) + Cl_2(g) \]Let the initial pressure of \(\displaystyle SO_2Cl_2\) be \(\displaystyle p_0\), and let \(\displaystyle x\) be the drop in its pressure by time \(\displaystyle t\). Since one mole of \(\displaystyle SO_2Cl_2\) gives one mole each of \(\displaystyle SO_2\) and \(\displaystyle Cl_2\):

	\(\displaystyle SO_2Cl_2\)	\(\displaystyle SO_2\)	\(\displaystyle Cl_2\)
at \(\displaystyle t=0\)	\(\displaystyle p_0\)	$\displaystyle 0$	$\displaystyle 0$
at \(\displaystyle t\)	\(\displaystyle p_0-x\)	\(\displaystyle x\)	\(\displaystyle x\)

The total pressure at time \(\displaystyle t\) is the sum of all three partial pressures: \[p_t = (p_0-x) + x + x = p_0 + x \quad\Rightarrow\quad x = p_t - p_0 \]So the partial pressure of the reactant that is actually left is\[p(SO_2Cl_2) = p_0 - x = p_0 - (p_t-p_0) = 2p_0 - p_t \]This is the step people skip — the total pressure given in the table is not the reactant's pressure; you have to subtract off what the products have contributed.Step $\displaystyle 1$: find \(\displaystyle k\) from the first data point.Here \(\displaystyle p_0 = 0.5\ \text{atm}\) (the pressure at \(\displaystyle t=0\), Experiment $\displaystyle 1$). For a first-order reaction, using pressure in place of concentration:\[k = \frac{2.303}{t}\log\frac{p_0}{p(SO_2Cl_2)} = \frac{2.303}{t}\log\frac{p_0}{2p_0-p_t} \]From Experiment $\displaystyle 2$, \(\displaystyle t = 100\ \text{s}\), \(\displaystyle p_t = 0.6\ \text{atm}\):\[p(SO_2Cl_2) = 2(0.5) - 0.6 = 0.4\ \text{atm} \]\[k = \frac{2.303}{100\ \text{s}}\log\frac{0.5}{0.4} = \frac{2.303}{100\ \text{s}}\log(1.25) = \frac{2.303 \times 0.0969}{100\ \text{s}} \]\[k = 2.232\times10^{-3}\ \text{s}^{-1} \]Step $\displaystyle 2$: find the reactant's pressure when the total pressure is $\displaystyle 0.65$ atm.Using the same relation \(\displaystyle p(SO_2Cl_2) = 2p_0 - p_t\) with \(\displaystyle p_t = 0.65\ \text{atm}\):\[p(SO_2Cl_2) = 2(0.5) - 0.65 = 0.35\ \text{atm} \]Step $\displaystyle 3$: get the rate from the first-order rate law.For a first-order reaction, \(\displaystyle \text{Rate} = k \times [\text{reactant}]\); with pressure standing in for concentration,\[\text{Rate} = k \times p(SO_2Cl_2) \]\[\text{Rate} = \left(2.232\times10^{-3}\ \text{s}^{-1}\right)\times\left(0.35\ \text{atm}\right) \]\[\text{Rate} = 7.81\times10^{-4}\ \text{atm}\ \text{s}^{-1} \]Because concentration here is expressed as a partial pressure (atm) rather than mol L⁻¹, the rate correctly carries units of atm s⁻¹, not mol L⁻¹ s⁻¹ — don't relabel the unit out of habit.Answer: Rate \(\displaystyle \approx 7.81\times10^{-4}\ \text{atm s}^{-1}\) (with \(\displaystyle k \approx 2.23\times10^{-3}\ \text{s}^{-1}\)).

The Arrhenius equation turns into a straight line only after you take logs — plot \(\displaystyle \ln k\) (y-axis) against \(\displaystyle 1/T\) (x-axis), and the slope gives \(\displaystyle E_a\), the intercept gives \(\displaystyle A\).Starting from the Arrhenius equation\[k = A\,e^{-E_a/RT} \]taking the natural log of both sides gives\[\ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T} \]This is a straight line \(\displaystyle y = mx + c\) with \(\displaystyle y=\ln k\), \(\displaystyle x = 1/T\), slope \(\displaystyle m = -E_a/R\), and intercept \(\displaystyle c = \ln A\). So a graph of \(\displaystyle \ln k\) vs \(\displaystyle 1/T\) lets you read off \(\displaystyle E_a\) and \(\displaystyle A\) without ever plugging numbers into the exponential form directly.Before plotting anything, fix the two traps in this data: temperature must be in kelvin (not °C), and the tabulated numbers are \(\displaystyle 10^{5}\times k\), not \(\displaystyle k\) itself.\(\displaystyle T/\mathrm{K} = T/^{\circ}\mathrm{C} + 273\), and \(\displaystyle k = (\text{tabulated value})\times10^{-5}\,\mathrm{s^{-1}}\). Working these out, and \(\displaystyle 1/T\) and \(\displaystyle \ln k\) alongside them:

\(\displaystyle T/^{\circ}\mathrm{C}\)	\(\displaystyle T/\mathrm{K}\)	\(\displaystyle k/\mathrm{s^{-1}}\)	\(\displaystyle 10^{3}\times(1/T)/\mathrm{K^{-1}}\)	\(\displaystyle \ln k\)
$\displaystyle 0$	$\displaystyle 273$	\(\displaystyle 7.87\times10^{-7}\)	$\displaystyle 3.663$	\(\displaystyle -14.055\)
$\displaystyle 20$	$\displaystyle 293$	\(\displaystyle 1.70\times10^{-5}\)	$\displaystyle 3.413$	\(\displaystyle -10.982\)
$\displaystyle 40$	$\displaystyle 313$	\(\displaystyle 2.57\times10^{-4}\)	$\displaystyle 3.195$	\(\displaystyle -8.266\)
$\displaystyle 60$	$\displaystyle 333$	\(\displaystyle 1.78\times10^{-3}\)	$\displaystyle 3.003$	\(\displaystyle -6.331\)
$\displaystyle 80$	$\displaystyle 353$	\(\displaystyle 2.14\times10^{-2}\)	$\displaystyle 2.833$	\(\displaystyle -3.844\)

Plotting these five \(\displaystyle (1/T,\ \ln k)\) points gives a straight line falling from upper left (low \(\displaystyle T\), very negative \(\displaystyle \ln k\)) to lower right (high \(\displaystyle T\), less negative \(\displaystyle \ln k\)) — a negative slope, exactly as the Arrhenius equation predicts, since \(\displaystyle k\) grows as \(\displaystyle T\) rises.The slope of the best-fit line through those points is what you're actually reading off the graph — here it's obtained from the least-squares slope formula,\[m = \frac{\sum_i (x_i-\bar x)(y_i-\bar y)}{\sum_i (x_i-\bar x)^2},\qquad x_i = \frac1{T_i},\ \ y_i=\ln k_i \]The means of the five points are \(\displaystyle \bar x = 3.2213\times10^{-3}\,\mathrm{K^{-1}}\) and \(\displaystyle \bar y = -8.6959\). Carrying these through the sums gives\[m = -1.2103\times10^{4}\ \mathrm{K} \]and the intercept\[c = \bar y - m\bar x = -8.6959 - (-1.2103\times10^{4})(3.2213\times10^{-3}) = 30.293 \]Reading \(\displaystyle E_a\) off the slope: \(\displaystyle m=-E_a/R\), so \(\displaystyle E_a=-mR\). With \(\displaystyle R = 8.314\ \mathrm{J\,mol^{-1}K^{-1}}\),\[E_a = -m R = (1.2103\times10^{4}\,\mathrm{K})(8.314\ \mathrm{J\,mol^{-1}K^{-1}}) = 1.006\times10^{5}\ \mathrm{J\,mol^{-1}} \]\[E_a \approx 101\ \mathrm{kJ\,mol^{-1}} \](This lands close to the accepted experimental activation energy for \(\displaystyle N_2O_5\) decomposition, which is a useful sanity check that the line was read correctly.)Reading \(\displaystyle A\) off the intercept: \(\displaystyle c = \ln A\), so \(\displaystyle A = e^{c}\).\[A = e^{30.293} = 1.43\times10^{13}\ \mathrm{s^{-1}} \]A pre-exponential factor of this order (\(\displaystyle \sim10^{13}\,\mathrm{s^{-1}}\)) is typical for a unimolecular gas-phase decomposition — it's roughly the frequency of a bond vibration, which is exactly the physical picture behind \(\displaystyle A\) for this kind of reaction.Predicting \(\displaystyle k\) at a temperature not in the table just means reading a point on the same line — plug \(\displaystyle T\) into the fitted equation, don't re-derive anything.\[\ln k = c + m\cdot\frac1T = 30.293 - (1.2103\times10^{4})\cdot\frac1T \]At \(\displaystyle 30^{\circ}\mathrm{C} = 303\ \mathrm{K}\): \(\displaystyle 1/T = 3.3003\times10^{-3}\,\mathrm{K^{-1}}\)\[\ln k = 30.293 - (1.2103\times10^{4})(3.3003\times10^{-3}) = -9.652 \] \[k = e^{-9.652} = 6.43\times10^{-5}\ \mathrm{s^{-1}} \]At \(\displaystyle 50^{\circ}\mathrm{C} = 323\ \mathrm{K}\): \(\displaystyle 1/T = 3.0960\times10^{-3}\,\mathrm{K^{-1}}\)\[\ln k = 30.293 - (1.2103\times10^{4})(3.0960\times10^{-3}) = -7.178 \] \[k = e^{-7.178} = 7.63\times10^{-4}\ \mathrm{s^{-1}} \]Both predictions sit exactly where they should: the value at \(\displaystyle 30^{\circ}\mathrm{C}\) falls between the measured \(\displaystyle k\) at \(\displaystyle 20^{\circ}\mathrm{C}\) (\(\displaystyle 1.70\times10^{-5}\)) and \(\displaystyle 40^{\circ}\mathrm{C}\) (\(\displaystyle 2.57\times10^{-4}\)), and the value at \(\displaystyle 50^{\circ}\mathrm{C}\) falls between \(\displaystyle 40^{\circ}\mathrm{C}\) (\(\displaystyle 2.57\times10^{-4}\)) and \(\displaystyle 60^{\circ}\mathrm{C}\) (\(\displaystyle 1.78\times10^{-3}\)) — confirming the line was fitted correctly.Answer: \(\displaystyle E_a \approx 101\ \mathrm{kJ\,mol^{-1}}\) and \(\displaystyle A \approx 1.43\times10^{13}\ \mathrm{s^{-1}}\); at \(\displaystyle 30^{\circ}\mathrm{C}\), \(\displaystyle k \approx 6.43\times10^{-5}\ \mathrm{s^{-1}}\), and at \(\displaystyle 50^{\circ}\mathrm{C}\), \(\displaystyle k \approx 7.63\times10^{-4}\ \mathrm{s^{-1}}\).

The Arrhenius equation is the bridge between the rate constant you measure and the pre-exponential factor you're asked for — solve it for \(\displaystyle A\) instead of \(\displaystyle k\).The Arrhenius equation is \[k = A\,e^{-E_a/RT} \] where \(\displaystyle k\) is the rate constant, \(\displaystyle A\) is the pre-exponential (frequency) factor, \(\displaystyle E_a\) is the activation energy, \(\displaystyle R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}\) is the gas constant, and \(\displaystyle T\) is the absolute temperature.Taking natural logs of both sides and switching to base-$\displaystyle 10$ logarithms (dividing by \(\displaystyle \ln 10 = 2.303\)) puts it in the form you actually calculate with: \[\ln k = \ln A - \frac{E_a}{RT} \quad\Longrightarrow\quad \log A = \log k + \frac{E_a}{2.303\,RT} \]The activation energy is given in kJ/mol but \(\displaystyle R\) is in J — convert before you divide, or the exponent is off by a factor of 1000. \[E_a = 179.9\ \text{kJ mol}^{-1} = 1.799\times10^{5}\ \text{J mol}^{-1} \]The remaining data: \(\displaystyle k = 2.418\times10^{-5}\ \text{s}^{-1}\), \(\displaystyle T = 546\ \text{K}\).Step $\displaystyle 1$ — evaluate \(\displaystyle 2.303RT\): \[2.303 \times 8.314\ \text{J K}^{-1}\text{mol}^{-1} \times 546\ \text{K} = 1.0454\times10^{4}\ \text{J mol}^{-1} \]Step $\displaystyle 2$ — evaluate \(\displaystyle E_a/(2.303RT)\): \[\frac{1.799\times10^{5}\ \text{J mol}^{-1}}{1.0454\times10^{4}\ \text{J mol}^{-1}} = 17.208 \] Notice the units of energy cancel top and bottom, leaving a pure number — exactly what a logarithm needs to be.Step $\displaystyle 3$ — evaluate \(\displaystyle \log k\): \[\log\left(2.418\times10^{-5}\right) = \log(2.418) + \log(10^{-5}) = 0.3835 - 5 = -4.6165 \]Step $\displaystyle 4$ — add them to get \(\displaystyle \log A\): \[\log A = -4.6165 + 17.208 = 12.592 \]Step $\displaystyle 5$ — undo the logarithm. Since \(\displaystyle A = 10^{\log A}\), \[A = 10^{12.592} = 10^{12}\times10^{0.592} = 10^{12}\times 3.905 = 3.905\times10^{12} \]\(\displaystyle A\) carries the same units as \(\displaystyle k\) (the Arrhenius equation only multiplies \(\displaystyle A\) by a dimensionless exponential), so here \(\displaystyle A\) is in \(\displaystyle \text{s}^{-1}\), the same as this first-order rate constant — not "per mole" or any other unit, since nothing in the exponential has units.Rounding to three significant figures, matching the precision of the given \(\displaystyle k\) and \(\displaystyle E_a\):Answer: \(\displaystyle A \approx 3.91\times10^{12}\ \text{s}^{-1}\)

For a first-order reaction, the concentration left after time \(\displaystyle t\) follows an exponential decay law — you cannot subtract the amount reacted linearly.The rate law for a first-order reaction \(\displaystyle A \to \text{Products}\) is\[\ln\frac{[A]_0}{[A]_t} = kt \]where \(\displaystyle [A]_0\) is the initial concentration, \(\displaystyle [A]_t\) is the concentration remaining at time \(\displaystyle t\), and \(\displaystyle k\) is the rate constant. This form comes directly from integrating \(\displaystyle -\dfrac{d[A]}{dt} = k[A]\), and it only holds because the reaction is first order — a key aside is that you must check the order before reaching for this formula, since a second-order reaction would need \(\displaystyle \dfrac{1}{[A]_t} - \dfrac{1}{[A]_0} = kt\) instead.Substituting the given valuesHere \(\displaystyle [A]_0 = 1.0\ \text{mol L}^{-1}\), \(\displaystyle k = 2.0 \times 10^{-2}\ \text{s}^{-1}\), and \(\displaystyle t = 100\ \text{s}\).\[\ln\frac{[A]_0}{[A]_t} = kt = (2.0 \times 10^{-2}\ \text{s}^{-1})(100\ \text{s}) = 2.0 \]Notice the units cancel cleanly: \(\displaystyle \text{s}^{-1} \times \text{s} = 1\) (dimensionless), which is exactly what a logarithm requires.Solving for \(\displaystyle [A]_t\)\[\frac{[A]_0}{[A]_t} = e^{2.0} \]\[[A]_t = [A]_0 \, e^{-2.0} = (1.0\ \text{mol L}^{-1}) \times e^{-2.0} \]Using \(\displaystyle e^{-2.0} = 0.13534\),\[[A]_t = 1.0\ \text{mol L}^{-1} \times 0.13534 = 0.13534\ \text{mol L}^{-1} \]A common slip here is to compute \(\displaystyle kt\) and then treat it as the fraction reacted rather than the exponent in \(\displaystyle e^{-kt}\) — the concentration remaining is the whole initial amount scaled by \(\displaystyle e^{-kt}\), not \(\displaystyle [A]_0(1-kt)\).Rounding to three significant figures (matching the two sig figs of \(\displaystyle k\) extended slightly for the exponential):\[[A]_t \approx 0.135\ \text{mol L}^{-1} \]Answer: The concentration of A remaining after $\displaystyle 100$ s is \(\displaystyle 0.135\ \text{mol L}^{-1}\) (obtained from \(\displaystyle [A]_t = [A]_0 e^{-kt} = 1.0 \times e^{-2.0}\)).

A first-order rate law means the fraction of reactant left depends only on \(\displaystyle kt\), never on the starting concentration — so you don't need to know how much sucrose you started with.Step $\displaystyle 1$: Get the rate constant from the half-life.For a first-order reaction the half-life is related to the rate constant \(\displaystyle k\) by \[t_{1/2} = \frac{0.693}{k} \] Rearranging for \(\displaystyle k\), with \(\displaystyle t_{1/2} = 3.00\) h: \[k = \frac{0.693}{t_{1/2}} = \frac{0.693}{3.00\ \text{h}} = 0.231\ \text{h}^{-1} \]Step $\displaystyle 2$: Use the first-order integrated rate law to relate concentration to time.The integrated first-order rate law is \[k = \frac{2.303}{t}\log\frac{[R]_0}{[R]} \] where \(\displaystyle [R]_0\) is the initial concentration of sucrose, \(\displaystyle [R]\) is the concentration left after time \(\displaystyle t\), and \(\displaystyle t\) is the elapsed time. Here \(\displaystyle t = 8\) h.A common slip: plugging \(\displaystyle t_{1/2}\) back into this formula instead of the actual elapsed time. The half-life was only used to find \(\displaystyle k\) in Step $\displaystyle 1$ — from here on, \(\displaystyle t = 8\) h is the number that matters.Substituting \(\displaystyle k = 0.231\ \text{h}^{-1}\) and \(\displaystyle t = 8\) h: \[0.231\ \text{h}^{-1} = \frac{2.303}{8\ \text{h}}\log\frac{[R]_0}{[R]} \]Step $\displaystyle 3$: Solve for the concentration ratio.\[\log\frac{[R]_0}{[R]} = \frac{0.231 \times 8}{2.303} = \frac{1.848}{2.303} = 0.8024 \]Taking the antilog: \[\frac{[R]_0}{[R]} = 10^{0.8024} = 6.346 \]Step $\displaystyle 4$: Invert to get the fraction remaining, not the fraction reacted.The question asks what fraction of sucrose remains, which is \(\displaystyle [R]/[R]_0\), the reciprocal of what Step $\displaystyle 3$ gives: \[\frac{[R]}{[R]_0} = \frac{1}{6.346} = 0.1576 \]Rounding to three significant figures (matching the precision of the given \(\displaystyle t_{1/2}\)): \[\frac{[R]}{[R]_0} \approx 0.158 \]So about $\displaystyle 15.8$% of the original sucrose is still present after $\displaystyle 8$ hours — consistent with the fact that $\displaystyle 8$ hours is a little under three half-lives ($\displaystyle 3$ × $\displaystyle 3.00$ h = $\displaystyle 9.00$ h), during which the fraction remaining would fall to \(\displaystyle (1/2)^3 = 0.125\); since $\displaystyle 8$ h is slightly less than $\displaystyle 9$ h, a fraction slightly above $\displaystyle 0.125$ is exactly what's expected.Answer: The fraction of sucrose remaining after $\displaystyle 8$ hours is about $\displaystyle 0.158$ ($\displaystyle 15.8$%) of the original amount.

Match the given rate law against the Arrhenius equation term by term — the number in the exponent IS \(\displaystyle E_a/R\), nothing more needs to be calculated separately.The Arrhenius equation is\[k = A\,e^{-E_a/RT} \]where \(\displaystyle A\) is the pre-exponential (frequency) factor, \(\displaystyle E_a\) is the activation energy, \(\displaystyle R\) is the gas constant, and \(\displaystyle T\) is the absolute temperature.The given equation is\[k = (4.5\times10^{11}\,\text{s}^{-1})\,e^{-28000\,\text{K}/T} \]Comparing the two exponents directly,\[\frac{E_a}{RT} = \frac{28000\ \text{K}}{T} \]The \(\displaystyle T\) on both sides cancels, leaving\[\frac{E_a}{R} = 28000\ \text{K} \]This is the step people rush past: the "$\displaystyle 28000$" is not \(\displaystyle E_a\) itself — it is \(\displaystyle E_a/R\), so you still have to multiply by \(\displaystyle R\) to get an energy.Taking \(\displaystyle R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}\),\[E_a = 28000\ \text{K} \times 8.314\ \text{J K}^{-1}\text{mol}^{-1} \]\[E_a = 232792\ \text{J mol}^{-1} \]Converting to kJ (dividing by $\displaystyle 1000$, since $\displaystyle 1$ kJ = $\displaystyle 1000$ J):\[E_a = 232.792\ \text{kJ mol}^{-1} \approx 232.79\ \text{kJ mol}^{-1} \]The pre-exponential factor \(\displaystyle 4.5\times10^{11}\,\text{s}^{-1}\) is simply \(\displaystyle A\) — it plays no role in finding \(\displaystyle E_a\); it would only be needed if you were asked to calculate \(\displaystyle k\) at a given \(\displaystyle T\).Answer: \(\displaystyle E_a = 232.79\ \text{kJ mol}^{-1}\)

The Arrhenius equation becomes a straight line once you take logarithms — matching the given equation to that line hands you \(\displaystyle E_a\) directly, and lets you solve backwards for \(\displaystyle T\) from any known \(\displaystyle k\).The Arrhenius equation is\[k = A\,e^{-E_a/RT} \]where \(\displaystyle k\) is the rate constant, \(\displaystyle A\) is the pre-exponential (frequency) factor, \(\displaystyle E_a\) is the activation energy, \(\displaystyle R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}\) is the gas constant, and \(\displaystyle T\) is the absolute temperature in kelvin. Taking \(\displaystyle \log_{10}\) of both sides converts the exponential into a straight line:\[\log k = \log A - \frac{E_a}{2.303\,R}\times\frac{1}{T} \]Compare this term by term with the equation given for \(\displaystyle H_2O_2\) decomposition:\[\log k = 14.34 - 1.25\times10^{4}\ \text{K}\times\frac{1}{T} \]Matching the two, the slope tells you the activation energy:\[\frac{E_a}{2.303\,R} = 1.25\times10^{4}\ \text{K} \]Step $\displaystyle 1$ — Find \(\displaystyle E_a\)\[E_a = 2.303\,R \times \left(1.25\times10^{4}\ \text{K}\right) \]\[E_a = 2.303 \times 8.314\ \text{J K}^{-1}\text{mol}^{-1} \times 1.25\times10^{4}\ \text{K} \]\[E_a = 19.147\ \text{J K}^{-1}\text{mol}^{-1} \times 1.25\times10^{4}\ \text{K} = 2.3934\times10^{5}\ \text{J mol}^{-1} \]\[E_a \approx 239.3\ \text{kJ mol}^{-1} \]Step $\displaystyle 2$ — Turn the half-period into a rate constant, in the SAME time unit the data equation usesThe pre-exponential term \(\displaystyle \log A = 14.34\) makes \(\displaystyle A \approx 2.2\times10^{14}\), which is a per-second frequency factor — the kind of value quoted for a unimolecular gas-phase-type decomposition. So the \(\displaystyle k\) in this equation is in \(\displaystyle \text{s}^{-1}\), not \(\displaystyle \text{min}^{-1}\). The half-period given is in minutes, so it must be converted to seconds before it can be plugged in — mixing the two time units here is the step that silently wrecks the answer.\[t_{1/2} = 256\ \text{min} \times 60\ \text{s min}^{-1} = 1.536\times10^{4}\ \text{s} \]For a first-order reaction, the half-life doesn't depend on concentration:\[t_{1/2} = \frac{0.693}{k} \quad\Rightarrow\quad k = \frac{0.693}{t_{1/2}} \]\[k = \frac{0.693}{1.536\times10^{4}\ \text{s}} = 4.512\times10^{-5}\ \text{s}^{-1} \]Step $\displaystyle 3$ — Feed this \(\displaystyle k\) back into the original equation and solve for \(\displaystyle T\)\[\log\!\left(4.512\times10^{-5}\right) = 14.34 - 1.25\times10^{4}\ \text{K}\times\frac{1}{T} \]\[-4.346 = 14.34 - \frac{1.25\times10^{4}\ \text{K}}{T} \]Rearranging to isolate the \(\displaystyle 1/T\) term:\[\frac{1.25\times10^{4}\ \text{K}}{T} = 14.34 - (-4.346) = 18.686 \]\[T = \frac{1.25\times10^{4}\ \text{K}}{18.686} = 668.96\ \text{K} \]Rounding to three significant figures (consistent with the precision of the given constants),\[T \approx 669\ \text{K} \]Answer: \(\displaystyle E_a \approx 239.3\ \text{kJ mol}^{-1}\); the half-period becomes $\displaystyle 256$ minutes at \(\displaystyle T \approx 669\ \text{K}\) ($\displaystyle 668.96$ K).

When k changes with temperature, use the two-point Arrhenius equation — it lets you find one unknown (here, \(\displaystyle T_2\)) without ever calculating the pre-exponential factor \(\displaystyle A\).The Arrhenius equation is \[k = A\,e^{-E_a/RT} \] where \(\displaystyle A\) is the frequency factor, \(\displaystyle E_a\) is the activation energy, \(\displaystyle R\) is the gas constant, and \(\displaystyle T\) is the absolute temperature. Taking this at two temperatures and dividing eliminates \(\displaystyle A\), giving the two-point form: \[\log\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \]List what's given, and convert everything to consistent units first — this is the step most solutions get wrong, using °C directly or leaving \(\displaystyle E_a\) in kJ.\[k_1 = 4.5\times10^{3}\ \text{s}^{-1}, \qquad T_1 = 10 + 273 = 283\ \text{K} \] \[k_2 = 1.5\times10^{4}\ \text{s}^{-1}, \qquad E_a = 60\ \text{kJ mol}^{-1} = 6.0\times10^{4}\ \text{J mol}^{-1} \] \[R = 8.314\ \text{J K}^{-1}\text{mol}^{-1} \]Compute the rate ratio. \[\frac{k_2}{k_1} = \frac{1.5\times10^{4}}{4.5\times10^{3}} = 3.333 \] \[\log(3.333) = 0.5229 \]Substitute into the Arrhenius expression and solve for \(\displaystyle T_2\). \[0.5229 = \frac{6.0\times10^{4}}{2.303 \times 8.314}\left(\frac{1}{283}-\frac{1}{T_2}\right) \]First evaluate the coefficient: \[\frac{E_a}{2.303\,R} = \frac{6.0\times10^{4}}{19.147} = 3133.6\ \text{K} \]So \[0.5229 = 3133.6\left(\frac{1}{283}-\frac{1}{T_2}\right) \] \[\frac{1}{283}-\frac{1}{T_2} = \frac{0.5229}{3133.6} = 1.6686\times10^{-4}\ \text{K}^{-1} \]Now isolate \(\displaystyle 1/T_2\): \[\frac{1}{T_2} = \frac{1}{283} - 1.6686\times10^{-4} = 3.5336\times10^{-3} - 0.16686\times10^{-3} = 3.3667\times10^{-3}\ \text{K}^{-1} \]Invert to get \(\displaystyle T_2\) — not \(\displaystyle 1/T_2\), which is the value people mistakenly report. \[T_2 = \frac{1}{3.3667\times10^{-3}} = 297.0\ \text{K} \]That's \(\displaystyle 297 - 273 = 24\,^\circ\text{C}\) — only a $\displaystyle 14$° rise, because the exponential dependence of \(\displaystyle k\) on \(\displaystyle T\) means even a small temperature increase multiplies the rate constant several-fold.Answer: \(\displaystyle T_2 \approx 297\ \text{K}\) (about \(\displaystyle 24\,^\circ\text{C}\))

For a first-order reaction, \(\displaystyle \mathrm{ k = \dfrac{2.303}{t}\log\dfrac{a}{a-x} ^{-}}\) and because the same time \(\displaystyle t \) is used for two different extents of reaction at two different temperatures, that \(\displaystyle t \) cancels out completely. You never need to know its actual value.Here \(\displaystyle k \) is the rate constant, \(\displaystyle t \) the elapsed time, \(\displaystyle a \) the initial amount taken as $\displaystyle 100$ (so percentages read off directly), and \(\displaystyle a-x \) the amount left unreacted.Step $\displaystyle 1$ — Write \(\displaystyle k \) at each temperature using the first-order law.At \(\displaystyle 298\ \text{K} \), $\displaystyle 10$% has reacted, so \(\displaystyle a=100 \), \(\displaystyle a-x = 90 \): \[k_{298} = \frac{2.303}{t}\log\frac{100}{90} \]At \(\displaystyle 308\ \text{K} \), the same \(\displaystyle t \) gives $\displaystyle 25$% completion, so \(\displaystyle a-x = 75 \): \[k_{308} = \frac{2.303}{t}\log\frac{100}{75} \]This pairing is the part to get right: $\displaystyle 10$% goes with the lower temperature and $\displaystyle 25$% with the higher one, because only that way can the same \(\displaystyle t \) produce both extents — a faster reaction (higher \(\displaystyle T \)) covers more ground in the same time.Step $\displaystyle 2$ — Divide the two equations; \(\displaystyle t \) and \(\displaystyle 2.303 \) cancel.\[\frac{k_{298}}{k_{308}} = \frac{\log(100/90)}{\log(100/75)} = \frac{0.04576}{0.12494} = 0.3662 \]so\[\frac{k_{308}}{k_{298}} = \frac{1}{0.3662} = 2.730 \]Step $\displaystyle 3$ — Feed this ratio into the two-temperature Arrhenius equation to get \(\displaystyle E_a \).\[\log\frac{k_{308}}{k_{298}} = \frac{E_a}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \]with \(\displaystyle R = 8.314\ \text{J mol}^{-1}\text{K}^{-1} \), \(\displaystyle T_1 = 298\ \text{K} \), \(\displaystyle T_2 = 308\ \text{K} \).\[\log(2.730) = 0.4362,\qquad \frac{1}{298}-\frac{1}{308} = \frac{308-298}{298\times308} = \frac{10}{91784} = 1.090\times10^{-4}\ \text{K}^{-1} \]Substituting:\[0.4362 = \frac{E_a}{2.303\times8.314}\times\left(1.090\times10^{-4}\right) \]\[E_a = \frac{0.4362\times2.303\times8.314}{1.090\times10^{-4}} = 76664\ \text{J mol}^{-1} \approx 76.7\ \text{kJ mol}^{-1} \]Step $\displaystyle 4$ — Use this \(\displaystyle E_a \) with the given \(\displaystyle A \) in the Arrhenius equation to find \(\displaystyle k \) at $\displaystyle 318$ K.\[\log k = \log A - \frac{E_a}{2.303RT} \]with \(\displaystyle A = 4\times10^{10}\ \text{s}^{-1} \) and \(\displaystyle T = 318\ \text{K} \):\[\log A = \log(4\times10^{10}) = 10.602 \]\[\frac{E_a}{2.303RT} = \frac{76664}{2.303\times8.314\times318} = \frac{76664}{6088.8} = 12.591 \]\[\log k_{318} = 10.602 - 12.591 = -1.989 \]\[k_{318} = 10^{-1.989} = 1.03\times10^{-2}\ \text{s}^{-1} \]Since this is a first-order reaction, \(\displaystyle k \) carries units of \(\displaystyle \text{s}^{-1} \) — the same units as \(\displaystyle A \) — with no concentration unit attached, unlike second-order rate constants.Answer: \(\displaystyle E_a \approx 76.7\ \text{kJ mol}^{-1} \), and \(\displaystyle k \) at $\displaystyle 318$ K \(\displaystyle \approx 1.03\times10^{-2}\ \text{s}^{-1} \).

The Arrhenius equation connects a change in rate constant to a change in temperature through the activation energy — this is the two-temperature (integrated) form, so you don't need the pre-exponential factor \(\displaystyle A\) at all.\[\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]where A step people rush past: "rate quadruples" means \(\displaystyle k_2/k_1 = 4\), not that the rate itself is $\displaystyle 4$ — for a reaction whose rate law has the form rate \(\displaystyle = k[\text{reactant}]\), the rate ratio and the \(\displaystyle k\) ratio are the same number, so this substitution is valid.Given: \[T_1 = 293\ \text{K}, \qquad T_2 = 313\ \text{K}, \qquad \frac{k_2}{k_1} = 4 \]Left side — evaluate the logarithm.\[\ln(4) = 1.3863 \]Right side — the temperature term. Keep this as an exact fraction rather than subtracting two rounded decimals, since \(\displaystyle 1/293\) and \(\displaystyle 1/313\) are both small numbers whose difference otherwise loses precision:\[\frac{1}{T_1} - \frac{1}{T_2} = \frac{T_2 - T_1}{T_1 T_2} = \frac{313 - 293}{293 \times 313} = \frac{20}{91709}\ \text{K}^{-1} = 2.1808 \times 10^{-4}\ \text{K}^{-1} \]Now solve for \(\displaystyle E_a\):\[E_a = \frac{R \ln(k_2/k_1)}{\dfrac{1}{T_1} - \dfrac{1}{T_2}} \]Substituting:\[E_a = \frac{(8.314\ \text{J K}^{-1}\text{mol}^{-1})(1.3863)}{2.1808 \times 10^{-4}\ \text{K}^{-1}} \]\[E_a = \frac{11.526\ \text{J K}^{-1}\text{mol}^{-1}}{2.1808 \times 10^{-4}\ \text{K}^{-1}} = 52850\ \text{J mol}^{-1} \]Converting to kJ (the unit activation energies are conventionally reported in) and rounding once, to three significant figures — matching the precision of the given temperatures:\[E_a = 52850\ \text{J mol}^{-1} = 52.9\ \text{kJ mol}^{-1} \]Answer: The activation energy of the reaction is \(\displaystyle E_a \approx 52.9\ \text{kJ mol}^{-1}\) (\(\displaystyle \approx 5.29 \times 10^{4}\ \text{J mol}^{-1}\)).