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Calculate the value of k for this reaction based on the following data collected in three different experiments:

$\begin{array} {cccc} \text{Experiment} & [A]\; \text{in M} & [B]\; \text{in M} & \text{Initial Rate in M min}^{-1}\\ 1 & 0.105 & 0.15 & 1.8 \times 10^{-5}\\ 2 & 0.105 & 0.30 & 7.1 \times 10^{-5}\\ 3 & 0.052 & 0.30 & 3.5 \times 10^{-5}\\ \end{array} $

$\begin{array}{1 1} 7.5 \times 10^{-3} \;M^{-2}min^{-1} \\ 3.75 \times 10^{-3} \;M^{-2}min^{-1} \\ 1.875 \times 10^{-3} \;M^{-2}min^{-1} \\ 11.25 \times 10^{-3} \;M^{-2}min^{-1}\end{array}$

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Answer: $ 7.5 \times 10^{-3} \;M^{-2}min^{-1}$
Given $\begin{array} {cccc} \text{Experiment} & [A]\; \text{in M} & [B]\; \text{in M} & \text{Initial Rate in M min}^{-1}\\ 1 & 0.105 & 0.15 & 1.8 \times 10^{-5}\\ 2 & 0.105 & 0.30 & 7.1 \times 10^{-5}\\ 3 & 0.052 & 0.30 & 3.5 \times 10^{-5}\\ \end{array} $
The rate of the reaction $ = k [A]^ x [B]^ y$
Comparing Experiments 3 and 2, With [B] constant, as [A] is doubled, rate is doubled. Therefore rate is first order with respect to [A] $\rightarrow$ x = 1
Compare experiments 1 and 2: With [A] constant, as [B] is doubled, rate increases by ~4 times. Therefore rate is second order with respect to [B] $\rightarrow $ y = 2
$\Rightarrow$ Rate $ = k[A][B]^2$
To calculate $k$, Insert data from any experiment into the rate equation:
$\Rightarrow 3.5 \times 10^{-5} = k \times 0.052 \times 0.30^2 \rightarrow k \approx 7.5 \times 10^{-3} \;M^{-2}min^{-1}$
answered Jul 25, 2014 by balaji.thirumalai
edited Jul 25, 2014 by balaji.thirumalai
 

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