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The following data was collected in three different experiments:

$\begin{array} {cccc} \text{Experiment} & [A]\; \text{in M} & [B]\; \text{in M} & \text{Initial Rate in Ms}^{-1}\\ 1 & 0.185 & 0.133 & 3.35 \times 10^{-4}\\ 2 & 0.185 & 0.266 & 1.35 \times 10^{-3}\\ 3 & 0.370 & 0.133 & 6.75 \times 10^{-4}\\ \end{array} $

Now a fourth experiment was conducted where $[A] = 0.370\;M,\; [B] = 0.266\; M$. What is the rate observed in this experiment?

$\begin{array}{1 1} 2.67 \times 10^{-3} Ms^{-1} \\ 1.33 \times 10^{-3} Ms^{-1} \\ 5.12 \times 10^{-3} Ms^{-1} \\ 3.99 \times 10^{-3} Ms^{-1}\end{array}$

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Answer: $2.67 \times 10^{-3} Ms^{-1}$
Given $\begin{array} {cccc} \text{Experiment} & [A] & [B] & \text{Initial Rate in Ms}^{-1}\\ 1 & 0.185 & 0.133 & 3.35 \times 10^{-4}\\ 2 & 0.185 & 0.266 & 1.35 \times 10^{-3}\\ 3 & 0.370 & 0.133 & 6.75 \times 10^{-4}\\ \end{array}$
The rate of the reaction $ = k [A]^ x [B]^ y$
Comparing Experiments 1 and 3, 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:
$\Rightarrow 3.35 \times 10^{-4} = k \times 0.185 \times 0.133^2 \rightarrow k = 0.102\;M^{-2}s^{-1}$
We can calculate the rate of the fourth experiment as follows: Rate $= k [A][B]^2 = 0.102 \times 0.370 \times 0.266^2 = 2.67 \times 10^{-3} Ms^{-1}$
answered Jul 24, 2014 by balaji.thirumalai
 

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