Chemical Kinetics - Result Question 81
80. Rate of reaction, $A+B \rightarrow$ products is given below as a function of different initial concentrations of $A$ and $B$
| $[\boldsymbol{A}] mol / L$ | $[\boldsymbol{B}] (mol / L)$ | Initial rate $\left(mol L^{-1} min^{-1}\right)$ |
|---|---|---|
| $0.01$ | $0.01$ | $0.005$ |
| $0.02$ | $0.01$ | $0.010$ |
| $0.01$ | $0.02$ | $0.005$ |
Determine the order of the reaction with respect to $A$ and $B$. What is the half-life of $A$ in the reaction?
$(1982,4 M)$
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Answer:
Correct Answer: 80. $(1.386 $ $min)$
Solution:
- Looking at the rate data of experiment number $1$ and $2$ indicates that rate is doubled on doubling concentration of $A$ while concentration of $B$ is constant. Therefore, order with respect to $A$ is $1$. Similarly, comparing data of experiment number $1$ and $3$ , doubling concentration of $B$, while concentration of $A$ is constant, has no effect on rate.
Therefore, order with respect to $B$ is zero.
$ \begin{aligned} \Rightarrow \quad & \text { Rate } =k[A] \\ \Rightarrow \quad & k =\frac{0.005}{0.010}=0.5 min^{-1}=\frac{0.693}{t _{1 / 2}} \\ \Rightarrow \quad & t _{1 / 2} =\frac{0.693}{0.5}=1.386 min \end{aligned} $
BETA
