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Standard XII

Chemistry

Factors Affecting Rate of Reaction

For the react...

Question

For the reaction
$A + B \to product$
rate law is
rate $= k [A]^{2} [B]$
where $k = 5 \times 10^{- 5} m o l^{- 2} L^{2} m i n^{- 1}$
Determine the time (in minutes) in which concentration of 'A' becomes half of its initial concentration. If initial concentration of A and B are $0.2 M$ and $2 \times 10^{3} M$ respectively.

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Solution

Since B is in excess, the rate law will become
Rate = $k^{'} [A]^{2}$ where $k^{'} = k [B]$
$\frac{1}{[A]_{t}} - \frac{1}{[A]_{0}} = k^{'} t \frac{1}{[A]_{2} k^{'}} = t_{\frac{1}{2}} t_{\frac{1}{2}} = \frac{1}{0.2 \times 5 \times 10^{- 5} \times 2 \times 10^{3}} = 50 m i n u t e s$

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For the reaction A+B→product rate law is rate =k[A]2[B] where k=5×10−5 mol−2 L2 min−1 Determine the time (in minutes) in which concentration of 'A' becomes half of its initial concentration. If initial concentration of A and B are 0.2 M and 2×103 M respectively.

Since B is in excess, the rate law will become Rate = k′[A]2 where k′=k[B] 1[A]t−1[A]0=k′t1[A]2k′=t12t12=10.2×5×10−5×2×103=50 minutes

Since B is in excess, the rate law will become
Rate = $k^{'} [A]^{2}$ where $k^{'} = k [B]$
$\frac{1}{[A]_{t}} - \frac{1}{[A]_{0}} = k^{'} t \frac{1}{[A]_{2} k^{'}} = t_{\frac{1}{2}} t_{\frac{1}{2}} = \frac{1}{0.2 \times 5 \times 10^{- 5} \times 2 \times 10^{3}} = 50 m i n u t e s$