Find the values of $$A, B$$  and $$C$$  in the following table for the reaction $$X + Y \to Z.$$   The reaction is of first order $$w.r.t$$  $$X$$ and zero order $$w.r.t.$$  $$Y.$$

Exp.	$$\left[ X \right]\left( {mol\,{L^{ - 1}}} \right)$$	$$\left[ Y \right]\left( {mol\,{L^{ - 1}}} \right)$$	Initial rate $$\left( {mol\,{L^{ - 1}}\,{s^{ - 1}}} \right)$$
1.	0.1	0.1	$$2 \times {10^{ - 2}}$$
2.	$$A$$	0.2	$$4 \times {10^{ - 2}}$$
3.	0.4	0.4	$$B$$
4.	$$C$$	0.2	$$2 \times {10^{ - 2}}$$

Solution :
$${\text{Rate}} = k\left[ X \right]{\left[ Y \right]^0}$$
Rate is independent of the cone. of $$Y$$ and it depends only on the conc. of $$X$$ and it is the first order reaction.
From exp. $$\left( 1 \right),2 \times {10^{ - 2}} = k\left( {0.1} \right)\,\,\,...\left( {\text{i}} \right)$$
From exp. $$\left( 2 \right),4 \times {10^{ - 2}} = k\left( A \right)\,\,\,...\left( {{\text{ii}}} \right)$$
Dividing (ii) and (i), $$\frac{{4 \times {{10}^{ - 2}}}}{{2 \times {{10}^{ - 2}}}} = \frac{{k\left( A \right)}}{{k\left( {0.1} \right)}} = \frac{A}{{0.1}}$$
$$ \Rightarrow 2 \times 0.1 = A \Rightarrow A = 0.2\,mol\,{L^{ - 1}}$$
From exp. $$\left( 3 \right),B = k\left( {0.4} \right)\,\,\,...\left( {{\text{iii}}} \right)$$
Dividing (iii) and (i), $$\frac{B}{{2 \times {{10}^{ - 2}}}} = \frac{{k\left( {0.4} \right)}}{{k\left( {0.1} \right)}} = 4$$
$$ \Rightarrow B = 4 \times 2 \times 10 - 2 = $$     $$8 \times {10^{ - 2}}\,mol\,{L^{ - 1}}\,{s^{ - 1}}$$
From exp. $$\left( 4 \right),2 \times {10^{ - 2}} = k\left( C \right)\,\,\,...\left( {{\text{iv}}} \right)$$
Dividing (iv) and (i), $$\frac{{2 \times {{10}^{ - 2}}}}{{2 \times {{10}^{ - 2}}}} = \frac{{k\left( C \right)}}{{k\left( {0.1} \right)}} = \frac{C}{{0.1}}$$
$$ \Rightarrow C = 0.1\,mol\,{L^{ - 1}}$$