Question

	$\left[S{O}_3\right]\left(mol/L\right)$	$\left[H_{2}O\right]\left(mol/L\right)$	Rate
Trial 1	0.1	0.01	0.013
Trial 2	0.2	0.01	0.052
Trial 3	X	0.02	0.234
Trial 4	0.1	0.03	0.039

Consider the following reaction and experimental data:
$S{O}_3+H_{2}O\to H_{2}S{O}_4$
(A) What is the value of $X$?
(B) What is the order of the reaction?
(C) What is the rate constant?
(D) What would be the rate if $\left[S{O}_3\right]$ in trial 4 were raised to $0.2$?

• A. A. $X=0.5$; B. $8$; C. $130$; D. $0.156$ units
• B. A. $X=0.3$; B. $3$; C. $130$; D. $0.156$ units
• C. A. $X=0.3$; B. $5$; C. $110$; D. $0.197$ units
• D. A. $X=0.7$; B. $9$; C. $110$; D. $0.156$ units

Answer 1

The correct option is B A. $X=0.3$; B. $3$; C. $130$; D. $0.156$ units

$Rate$ = $K{\left[A\right]}^x{\left[B\right]}^y$
$0.013$ = $K\left(0.1{)}^x\right(0.01{)}^y$ ..........(i)
$0.052$ = $K\left(0.2{)}^x\right(0.01{)}^y$ ..........(ii)
$0.234$ = $K\left(X{)}^x\right(0.02{)}^y$ ...........(iii)
$0.039$ = $K\left(0.1{)}^x\right(0.03{)}^y$ ..........(iv)
now,divide eq. (ii) from eq. (i) ,
$\frac{0.052}{0.013}$ = $\frac{K\left(0.2{)}^x\right(0.01{)}^y}{K\left(0.1{)}^x\right(0.01{)}^y}$
$\left(4\right)$ = $(2{)}^x$
$(2{)}^2$ = $(2{)}^x$
$x=2$
similarly,divide eq. (iv) from eq. (i) ,
$\frac{0.039}{0.013}$ = $\frac{K\left(0.1{)}^x\right(0.03{)}^y}{K\left(0.1{)}^x\right(0.01{)}^y}$
$(3{)}^1$ = $(3{)}^y$
$y=1$
put the values of x and y in eq. (i),we get,
$0.013$ = K $\left(0.1{)}^2\right(0.01{)}^1$
$K$ = $\frac{0.013}{0.1\times 0.1\times 0.01}$
$K=130$
Now,from the values of x,y and K,in eq. (iii),we get ,
$0.234$ = K$\left(X{)}^x\right(0.02{)}^y$
$0.234$ = $130\times (X{)}^2\times (0.02{)}^1$
$(X{)}^2$ = $\frac{0.234}{130\times 0.02}$
$(X{)}^2$ = $\frac{234}{130\times 2\times 10}$
$(X{)}^2$ = $\frac{117}{1300}$
$(X{)}^2$ = $0.09$
$(X{)}^2$ = $(0.3{)}^2$
$X=0.3$

According to question,if the value of $\left[S{O}_3\right]$ in trial 4 is $0.2$,then,

$Rate$ = $\left(130\right)\left(0.2{)}^2\right(0.03{)}^1$
$Rate$ = $130\times 0.2\times 0.2\times 0.03$
$Rate$ = $0.156$

hence, the answer is -
(B) A. $X=0.3$; B. $3$; C. $130$; D. $0.156$ units

Hence, te correct option is $\text{B}$