CBSE 2024 · Region 1 · Set 1 · Q23 · 3 marks
The following initial rate data were obtained for the reaction : \[2 \mathrm{NO}(\mathrm{~g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NOBr}(\mathrm{~g}) \] Expt. No. $\displaystyle [\mathrm{NO}] / \mathrm{mol} \mathrm{L}^{-1}$ $\displaystyle \left[\mathrm{Br}_{2}\right] / \mathrm{mol} \mathrm{L}^{-1}$ Initial Rate $\displaystyle \left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)$ $\displaystyle 1$ $\displaystyle 0.05$ $\displaystyle 0.05$ $\displaystyle 1 \cdot 0 \times 10^{-3}$ $\displaystyle 2$ $\displaystyle 0.05$ $\displaystyle 0.15$ $\displaystyle 3 \cdot 0 \times 10^{-3}$ $\displaystyle 3$ $\displaystyle 0.15$ $\displaystyle 0.05$ $\displaystyle 9 \cdot 0 \times 10^{-3}$
(a)What is the order with respect to NO and $\displaystyle \mathrm{Br}_{2}$ in the reaction ?(b)Calculate the rate constant (k).(c)Determine the rate of reaction when concentration of NO and $\displaystyle \mathrm{Br}_{2}$ are $\displaystyle 0.4$ M and $\displaystyle 0.2$ M, respectively.
The following initial rate data were obtained for the reaction : \[2 \mathrm{NO}(\mathrm{~g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NOBr}(\mathrm{~g}) \]
| Expt. No. | $\displaystyle [\mathrm{NO}] / \mathrm{mol} \mathrm{L}^{-1}$ | $\displaystyle \left[\mathrm{Br}_{2}\right] / \mathrm{mol} \mathrm{L}^{-1}$ | Initial Rate $\displaystyle \left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)$ |
| $\displaystyle 1$ | $\displaystyle 0.05$ | $\displaystyle 0.05$ | $\displaystyle 1 \cdot 0 \times 10^{-3}$ |
| $\displaystyle 2$ | $\displaystyle 0.05$ | $\displaystyle 0.15$ | $\displaystyle 3 \cdot 0 \times 10^{-3}$ |
| $\displaystyle 3$ | $\displaystyle 0.15$ | $\displaystyle 0.05$ | $\displaystyle 9 \cdot 0 \times 10^{-3}$ |
(a)
What is the order with respect to NO and $\displaystyle \mathrm{Br}_{2}$ in the reaction ?
(b)
Calculate the rate constant (k).
(c)
Determine the rate of reaction when concentration of NO and $\displaystyle \mathrm{Br}_{2}$ are $\displaystyle 0.4$ M and $\displaystyle 0.2$ M, respectively.
Marking-scheme solution
\[\begin{aligned}
& \text { Rate }=\mathrm{k}[\mathrm{NO}]^{\mathrm{p}}\left[\mathrm{Br}_{2}\right]^{\mathrm{q}} \\
& 1.0 \times 10^{-3}=\mathrm{k}[0.05]^{\mathrm{p}}[0.05]^{\mathrm{q}} \\
& 3.0 \times 10^{-3}=\mathrm{k}[0.05]^{\mathrm{p}}[0.15]^{\mathrm{q}} \\
& 9.0 \times 10^{-3}=\mathrm{k}[0.15]^{\mathrm{p}}[0.05]^{\mathrm{q}}
\end{aligned}
\]
On Comparing (eq1) and (eq2)
\[\begin{aligned}
\left(\frac{1}{3}\right)= & \left(\frac{1}{3}\right)^{\mathrm{q}} \\
& \mathrm{q}=1 \\
& (\mathrm{eq} 1) \div(\mathrm{eq} 3) \\
& \left(\frac{1}{9}\right)=\left(\frac{1}{3}\right)^{\mathrm{p}} \\
& \mathrm{p}=2
\end{aligned}
\]
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