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Chemistry · 2022 · 5 marks
CBSE 2022 · Region 4 · Set 1 · Q12
Read the passage given below and answer the questions that follow: The rate law for a chemical reaction relates the reaction rate with the…
Read the passage given below and answer the questions that follow:
The rate law for a chemical reaction relates the reaction rate with the concentrations or partial pressures of the reactants. For a general reaction aA + bB → C with no intermediate steps in its reaction mechanism, meaning that it is an elementary reaction, the rate law is given by $\displaystyle \mathrm{r}=\mathrm{k}[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}$, where [A] and [B] express the concentrations of A and B in moles per litre. Exponents x and y vary for each reaction and are determined experimentally. The value of k varies with conditions that affect reaction rate, such as temperature, pressure, surface area, etc. The sum of these exponents is known as overall reaction order. A zero order reaction has a constant rate that is independent of the concentration of the reactants. A first order reaction depends on the concentration of only one reactant. A reaction is said to be second order when the overall order is two. Once we have determined the order of the reaction, we can go back and plug in one set of our initial values and solve for k.
(i)
Calculate the overall order of a reaction which has the following rate expression : \[\text { Rate }=\mathrm{k}[\mathrm{~A}]^{1 / 2}[\mathrm{~B}]^{3 / 2} \]
(ii)
What is the effect of temperature on rate of reaction ?
(iii)
What is meant by rate of a reaction?
(iv)
(a)
A first order reaction takes $\displaystyle 77 \cdot 78$ minutes for $\displaystyle 50 \%$ completion. Calculate the time required for $\displaystyle 30 \%$ completion of this reaction. $\displaystyle (\log 10=1, \log 7=0 \cdot 8450)$
OR
(b)
A first order reaction has a rate constant $\displaystyle 1 \times 10^{-3}$ per sec. How long will $\displaystyle 5$ g of this reactant take to reduce to $\displaystyle 3$ g ? \[(\log 3=0 \cdot 4771 ; \log 5=0 \cdot 6990) \]
Marking-scheme solution
(i) Order \(\displaystyle =\frac{1}{2}+\frac{3}{2}=2\)
(ii) The rate of a reaction increases with the increase in temperature.
(iii) The change in concentration of the reactants or products per unit time.
(iv)
\[\begin{aligned} & \mathrm{k}=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{77.78}=.008909 \mathrm{~min}^{-1} \\
& t=\frac{2.303}{\mathrm{k}} \log \frac{a}{a-x} \\
& t=\frac{2.303}{.008909} \log \frac{100}{100-30} \\
& t=258.5026(\log 10-\log 7) \\
& =258.5026 \times 0.155 \\
& =40.06 \mathrm{~min} \text { or } 40.02 \mathrm{~min}
\end{aligned}
\]
\[\begin{aligned}
& \mathrm{k}=\frac{2.303}{t} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]} \\
& t=\frac{2.303}{1 \times 10^{-3}} \log \frac{5}{3} \\
& t=2.303 \times 10^{3}[0.699-0.4771] \\
& t=511 \mathrm{~s}
\end{aligned}\]
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CBSE Class 12 Chemistry past-paper question from the 2022 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.