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	$2\times 10^{-2}$
2.	$A$	0.2	$4\times 10^{-2}$
3.	0.4	0.4	B
4.	C	0.2	$2\times 10^{-2}$

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

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. $\left(1\right),2\times 10^{-2}=k\left(0.1\right)\dots \left(i\right)$
From exp. $\left(2\right),4\times 10^{-2}=k\left(A\right)\dots \left(ii\right)$
Dividing $\left(ii\right)$ and $\left(i\right)$, $\frac{4\times 10^{-2}}{2\times 10^{-2}}=\frac{k\left(A\right)}{k\left(0.1\right)}=\frac{A}{0.1}$
$\Rightarrow 2\times 0.1=A$
$\Rightarrow A=0.2mol{L}^{-1}$
From exp. $\left(3\right)$, $B=k\left(0.4\right)\dots \left(iii\right)$
Dividing $\left(iii\right)$ and $\left(i\right)$, $\frac{B}{2\times 10^{-2}}=\frac{k\left(0.4\right)}{k\left(0.1\right)}=4$
$\Rightarrow B=4\times 2\times 10^{-2}=8\times 10^{-2}mol{L}^{-1}{s}^{-1}$
From exp. $\left(4\right),2\times 10^{-2}=k\left(C\right)\dots \left(iv\right)$
Dividing $\left(iv\right)$ and $\left(i\right)$, $\frac{2\times 10^{-2}}{2\times 10^{-2}}=\frac{k\left(C\right)}{k\left(0.1\right)}=\frac{C}{0.1}$
$\Rightarrow C=0.1mol{L}^{-1}$