Solveeit Logo
Login

Question

Question: For the non-stoichiometry reaction \(2A+B\to C+D\) , the following kinetic data were obtained in thr...

For the non-stoichiometry reaction 2A+B→C+D2A+B\to C+D , the following kinetic data were obtained in three separate experiments, all at 298 K. The rate law for the formation of C is:

Initial\text{ }Concentration\left( A \right) & Initial\text{ }Concentration\left( B \right) & Initial\text{ }rate\text{ }of\text{ }formation\text{ }of\text{ }C\left( mol{{L}^{-1}}{{S}^{-1}} \right) \\\ 0.1M & 0.1M & 1.2\times {{10}^{-3}} \\\ 0.1M & 0.2M & 1.2\times {{10}^{-3}} \\\ 0.2M & 0.1M & 2.4\times {{10}^{-3}} \\\ \end{matrix}$$ (A) $\dfrac{dc}{dt}=k\left[ A \right]{{\left[ B \right]}^{2}}$ (B) $\dfrac{dc}{dt}=k\left[ A \right]$ (C) $\dfrac{dc}{dt}=k\left[ A \right]\left[ B \right]$ (D) $\dfrac{dc}{dt}=k{{\left[ A \right]}^{2}}\left[ B \right]$
Explanation

Solution

The rate law can be experimentally determined and also be used to predict the relationship between the rate of a reaction and the concentrations of reactants. We are going to find the relationship between the given conditions and through that we can derive rate law for the formation of C.

Complete step by step answer:
- Let’s start with the concept of rate law. It is an equation which relates the reaction rate with the partial pressures or concentrations of the reactants.
The rate law equation can be written as follows,
dcdt=k[A]x[B]y\dfrac{dc}{dt}=k{{\left[ A \right]}^{x}}{{\left[ B \right]}^{y}}
The dcdt\dfrac{dc}{dt} for C at various concentrations are
1.2×10−3=k(0.1)x(0.1)y1.2\times 1{{0}^{-3}}=k{{\left( 0.1 \right)}^{x}}{{\left( 0.1 \right)}^{y}} ⋯⋯(i)
1.2×10−3=k(0.1)x(0.2)y1.2\times {{10}^{-3}}=k{{\left( 0.1 \right)}^{x}}{{\left( 0.2 \right)}^{y}} ⋯⋯(ii)
2.4×10−3=k(0.2)x(0.1)y2.4\times {{10}^{-3}}=k{{\left( 0.2 \right)}^{x}}{{\left( 0.1 \right)}^{y}} ⋯⋯(iii)
In order to solve x divide the equation (iii) by equation (i) we get,
2.4×10−31.2×10−3=k(0.2)x(0.1)yk(0.1)x(0.1)y\dfrac{2.4\times {{10}^{-3}}}{1.2\times {{10}^{-3}}}=\dfrac{k{{\left( 0.2 \right)}^{x}}{{\left( 0.1 \right)}^{y}}}{k{{\left( 0.1 \right)}^{x}}{{\left( 0.1 \right)}^{y}}}
Which on solving becomes
2={{\left( 2 \right)}^{x}}$$$$$$ x=1Nowweneedtofindthevalueofy.Forthatdividetheequation(iii)byequation(ii) Now we need to find the value of y.For that divide the equation (iii) by equation (ii) \dfrac{2.4\times {{10}^{-3}}}{1.2\times {{10}^{-3}}}=\dfrac{k{{\left( 0.2 \right)}^{x}}{{\left( 0.1 \right)}^{y}}}{k{{\left( 0.1 \right)}^{x}}{{\left( 0.2 \right)}^{y}}} 2={{\left( 2 \right)}^{x}}\left( \dfrac{0.1}{0.2} \right)Weknowthatvalueofxis1.Substitutethisintheaboveequation We know that value of x is 1. Substitute this in the above equation 2={{\left( 2 \right)}^{1}}{{\left( \dfrac{0.1}{0.2} \right)}^{y}} 1={{\left( 0.5 \right)}^{y}} {{1}^{0}}={{\left( 0.5 \right)}^{y}} y=0$$

Thus we can substitute this values of x and y in the rate law and we will get as follows
dcdt=k[A]1[B]0\dfrac{dc}{dt}=k{{\left[ A \right]}^{1}}{{\left[ B \right]}^{0}}
dcdt=k[A]\dfrac{dc}{dt}=k\left[ A \right]
Therefore the answer is option (B) dcdt=k[A]\dfrac{dc}{dt}=k\left[ A \right]
So, the correct answer is “Option C”.

Note: The answer can be found without these calculations too. From the first and second experiment, it is seen that when the initial concentration of A is kept constant and the initial concentration of B is doubled, the rate of the formation of C will remain unaffected. That is, the rate of formation of C is independent of the initial concentration of B. Also, from the first and third experiment, it can be observed that when the initial concentration of B is kept constant and the initial concentration of A is doubled, the rate of formation of C is doubled. That is, the rate of formation of C will be directly proportional to the initial concentration of A. From this we can conclude that, the rate of formation of A can be given by the expression dcdt=k[A]\dfrac{dc}{dt}=k\left[ A \right]