Question

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.	$[X] (mol\, L^{-1})$	$[Y](mol\, L^{-1})$	Initial rate $(mol \, L^{-1} s^{-1})$
1	0.1	0.1	$2 \times10^{-2}$
2.	$A$	0.2	$4 \times10^{-2}$
3.	0.4	0.4	B
4.	C	0.2	$2\times 10^{-2}$

• A. $A = 0.2\, mol\, L^{-1}, B = 8 \times 10^{-2}\, mol \,L^{-1} s^{-1}, C = 0.1 mol \,L^{-1}$
• B. $A = 0.4 \,mol\, L^{-1}, B = 4 \times 10^{-2} \,mol\, L^{-1} s^{-1} , C = 0.2 mol \,L^{-1}$
• C. $A = 0.2\, mol\, L^{-1}, B = 2 \times 10^{-2} \,mol\, L^{-1} s^{-1}, C = 0.4 \,mol \,L^{-1}$
• D. $A = 0.4 \,mol\, L^{-1}, B = 2\times 10^{-2}\, mol\, L^{-1} s^{-1}, C=0.4\, mol\, L^{-1}$

Answer 1

Answer: (A)

Rate$ = k\left[X\right]\left[Y\right]^{0}$
Rate is independent of the conc, of $Y$ and it depends onlyon the conc, of $X$ and it is the first order reaction.
From exp. $\left(1\right), 2\times10^{-2}=k\left(0.1\right) \ldots\left(i\right)$
From exp. $\left(2\right),4\times10^{-2}=k\left(A\right) \ldots\left(ii\right)$
Dividing $\left(ii\right)$ and $\left(i\right)$, $\frac{4\times10^{-2}}{2\times10^{-2}}=\frac{k\left(A\right)}{k\left(0.1\right)}=\frac{A}{0.1}$
$\Rightarrow 2\times0.1=A$
$\Rightarrow A=0.2 mol \,L^{-1}$
From exp. $\left(3\right)$, $B = k\left(0.4\right) \ldots\left(iii\right)$
Dividing $\left(iii\right)$ and $\left(i\right)$, $\frac{B}{2\times10^{-2}}=\frac{k\left(0.4\right)}{k\left(0.1\right)}=4$
$\Rightarrow B=4\times2\times10^{-2}=8\times10^{-2}mol\,L^{-1}s^{-1}$
From exp. $\left(4\right), 2\times10^{-2}=k\left(C\right) \ldots\left(iv\right)$
Dividing $\left(iv\right)$ and $\left(i\right)$, $\frac{2\times10^{-2}}{2\times10^{-2}}=\frac{k\left(C\right)}{k\left(0.1\right)}=\frac{C}{0.1}$
$\Rightarrow C=0.1 \, mol\, L^{-1}$