Answer
Verified
40.8k+ views
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.
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.
Recently Updated Pages
silver wire has diameter 04mm and resistivity 16 times class 12 physics JEE_Main
A parallel plate capacitor has a capacitance C When class 12 physics JEE_Main
Let gx 1 + x x and fx left beginarray20c 1x 0 0x 0 class 12 maths JEE_Main
A series combination of n1 capacitors each of value class 12 physics JEE_Main
When propyne is treated with aqueous H2SO4 in presence class 12 chemistry JEE_Main
Which of the following is not true in the case of motion class 12 physics JEE_Main
Other Pages
The radius of the sphere is 43 pm 01cm The percentage class 11 physics JEE_Main
The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Which halogen in its reference state will have the class 11 chemistry JEE_Main
Explain the construction and working of a GeigerMuller class 12 physics JEE_Main
when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main