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Question: The reaction between A and B is first order with respect to A and zero order with respect to B. Fill...

The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:

Experiment[A]/molL−1\left[ A \right]/mol{{L}^{-1}}[B]/molL−1\left[ B \right]/mol{{L}^{-1}}initial rate/molL−1min⁡−1initial~rate/mol{{L}^{-1}}{{\min }^{-1}}
10.10.12.0×10−22.0\times {{10}^{-2}}
2-0.24.0×10−24.0\times {{10}^{-2}}
30.40.4
4-0.22.0×10−22.0\times {{10}^{-2}}
Explanation

Solution

When a chemical reaction takes place, it is important to know that by which mechanism and at which rate it has taken place, hence, the Law of Mass Action states that the rate depends on the concentration of the reactants and products which takes place at constant temperature.
Rate=k[A] [B], where k is the rate constant.

Complete Solution :
In this question, four experiments are conducted and their data is given.
Since, the A is first order reaction and B is zero order, the rate equation becomes,
⇒Rate=k[A]1[B]0\Rightarrow Rate=k{{[A]}^{1}}{{[B]}^{0}}
⇒Rate=k[A]\Rightarrow Rate=k[A]

  1. In the first experiment:
    ⇒2.0×10−2molL−1min⁡−1=k[0.1molL−1]\Rightarrow 2.0\times {{10}^{-2}}mol{{L}^{-1}}{{\min }^{-1}}=k[0.1mol{{L}^{-1}}]
    ⇒k=0.2min⁡−1\Rightarrow k=0.2{{\min }^{-1}}

  2. In the second experiment:
    ⇒Rate=k[A]\Rightarrow Rate=k[A]
    ⇒[A]=ratek\Rightarrow [A]=\dfrac{rate}{k}
    ⇒[A]=4×10−2molL−1min⁡−10.2min⁡−1\Rightarrow [A]=\dfrac{4\times {{10}^{-2}}mol{{L}^{-1}}{{\min }^{-1}}}{0.2{{\min }^{-1}}}
    [A]=0.20M

  3. In third experiment:
    ⇒Rate=k[A]\Rightarrow Rate=k[A]
    ⇒0.2min⁡−1×0.4M\Rightarrow 0.2{{\min }^{-1}}\times 0.4M
    ⇒0.8min⁡−1\Rightarrow 0.8{{\min }^{-1}}
    ⇒8×10−2Mmin⁡−1\Rightarrow 8\times {{10}^{-2}}M{{\min }^{-1}}

  4. In fourth experiment,
    ⇒Rate=k[A]\Rightarrow Rate=k[A]
    ⇒[A]=ratek\Rightarrow [A]=\dfrac{rate}{k}
    ⇒2.0×10−2Mmin⁡−10.2min⁡−1\Rightarrow \dfrac{2.0\times {{10}^{-2}}M{{\min }^{-1}}}{0.2{{\min }^{-1}}}
    =0.1M
    Hence, the table will be filled as:

Experiment[A]/molL−1\left[ A \right]/mol{{L}^{-1}}[B]/molL−1\left[ B \right]/mol{{L}^{-1}}initia lrate/molL−1min⁡−1initia~lrate/mol{{L}^{-1}}{{\min }^{-1}}
10.10.12.0×10−22.0\times {{10}^{-2}}
20.20.24.0×10−24.0\times {{10}^{-2}}
30.40.48×10−28\times {{10}^{-2}}
40.10.22.0×10−22.0\times {{10}^{-2}}

Note: Order of the reaction is the reaction is dependent on the power of the reactants involved in the reaction. It can be obtained by adding the powers of reactants involved in the reaction. The value of the order can be an integer or in fraction or zero. We have not taken the concentration of reactant B here because the order of the B is zero and when the power of a variable is zero it is considered as 1.