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# The reaction $2{I^ - }_{(aq)} + {S_2}{O_8}_{(aq)}^{2 - } \to {I_{2(aq)}} + 2SO_{4(aq)}^{2 - }$ was studied at 25 $^\circ C$. the following results were obtained. Where, Rate = $- \dfrac{{d[{S_2}{O_8}_{}^{2 - }]}}{{dt}}$[${I^ - }$]$[{S_2}{O_8}_{}^{2 - }]$Initial rate($mol{L^{ - 1}}{S^{ - 1}}$)0.080.040 $12.5 \times {10^{ - 6}}$0.040.040$6.25 \times {10^{ - 6}}$0.080.020$5.56 \times {10^{ - 6}}$0.0320.040$4.35 \times {10^{ - 6}}$0.0600.030$6.42 \times {10^{ - 6}}$The value of the rate constant is:$A.\;\;\;\;\;1.8 \times {10^{ - 3}}Lmo{l^{ - 1}}{s^{ - 1}} \\ B.\;\;\;\;\;0.9 \times {10^{ - 3}}Lmo{l^{ - 1}}{s^{ - 1}} \\ C.\;\;\;\;\;3.7 \times {10^{ - 3}}Lmo{l^{ - 1}}{s^{ - 1}} \\ D.\;\;\;\;\;4.5 \times {10^{ - 3}}Lmo{l^{ - 1}}{s^{ - 1}} \\$

Last updated date: 16th Sep 2024
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Hint: In order to find the rate constant, we’ll need to understand the process of calculating the rate constant. The rate constant of any given chemical reaction is dependent on the 2 main factors: 1. The initial rate of the reaction 2. The order of the reaction with respect to the reactants involved.

Complete Step-by-Step Solution:
In order to calculate the order of the reaction with respect to the reactants involved:
For [${I^ - }$]:
Comparing reactions 01 and 02, we can see that the concentration of $[{S_2}{O_8}_{}^{2 - }]$is kept constant. On the other hand, the concentration value of [${I^ - }$] has been halved. In correspondence, the rate of the reaction changes from $12.56 \times {10^{ - 6}}$ in reaction 01 to $6.25 \times {10^{ - 6}}$in reaction 02. This is a decrease of $6.25 \times {10^{ - 6}}$in the value of the rate of the reaction, i.e. the value of the reaction rate also halved. Since the concentration of [${I^ - }$] was halved to achieve the new reaction rate, the order of reaction with respect to [${I^ - }$]is 1.

For $[{S_2}{O_8}_{}^{2 - }]$:
Comparing reactions 01 and 03, we can see that the concentration [${I^ - }$] of is kept constant. On the other hand, the concentration value of $[{S_2}{O_8}_{}^{2 - }]$has been halved. In correspondence, the rate of the reaction changes from $12.56 \times {10^{ - 6}}$ in reaction 01 to $5.56 \times {10^{ - 6}}$ in reaction 02. This is a decrease of 2.248 times the original value, in the value of the rate of the reaction. Since the concentration of $[{S_2}{O_8}_{}^{2 - }]$ was halved to achieve the new reaction rate, the order of reaction with respect to $[{S_2}{O_8}_{}^{2 - }]$is $\dfrac{{2.248}}{2} = 1.124 \approx 1$
Hence, the expression for the rate of the reaction can be represented as:
$R = k{[{I^ - }]^m}{[{S_2}{O_8}_{}^{2 - }]^n}$
Where, R is the rate of the reaction
k is the rate constant
m is the order of the reaction for$[{I^ - }] = 1$
n is the order of the reaction for $[{S_2}{O_8}_{}^{2 - }] = 1$
Hence substituting the values obtained and from reaction 01, we get,
$k = \dfrac{R}{{{{[{I^ - }]}^m}{{[{S_2}{O_8}_{}^{2 - }]}^n}}}$
$k = \dfrac{{12.56 \times {{10}^{ - 6}}}}{{{{[{I^ - }]}^1}{{[{S_2}{O_8}_{}^{2 - }]}^1}}}$
$k = \dfrac{{12.56 \times {{10}^{ - 6}}}}{{{{[0.08]}^1}{{[0.04]}^1}}}$
$k = 3.7 \times {10^{ - 3}}$$Lmo{l^{ - 1}}{s^{ - 1}}$

Hence, Option C is the correct option.

Note: In order to determine the reaction order from experimental data either the differential rate law or the integrated rate law can be used. Often, the values of the exponents used in the rate law are the positive integers: 1 and 2 or even 0. Hence, the reactions are zeroth, first, or second order in each reactant.