Question

The following information is available for the reaction:
$R\to P;rate=r_1$
$R\stackrel{A}{\to }P;rate=r_2$
$R\stackrel{B}{\to }P;rate=r_3$
$R\stackrel{C}{\to }P;rate=r_4$
$R\stackrel{D}{\to }P;rate=r_5$
$R\stackrel{A+C}{\to }P;rate=r_6$
$R\stackrel{A+D}{\to }P;rate=r_7$
If $r_3<r_1=r_4=r_5<r_7<r_2<r_6,$ then the only incorrect statement is

• A. B is an inhibitor.
• B. C is a catalytic promoter of catalyst A.
• C. D is catalytic poison of catalyst A.
• D. A is catalytic promoter of catalyst C.

Answer 1

The correct option is D A is catalytic promoter of catalyst C.

Given,
$R\to P;rate=r_1$
and $r_3<r_1=r_4=r_5<r_7<r_2<r_6$
(a)
$R\to P$
The rate of the reaction is $r_1$
For,
$R\stackrel{B}{\to }P$
$Rate=r_3$
Given, $r_3<r_1$
So, B is inhibiting the reaction i.e., B is an inhibitor


(b)

$R\stackrel{A}{\to }P$ rate = $r_2$
$R\stackrel{C}{\to }P$ rate = $r_4$
$R\stackrel{A+C}{\to }P$ rate = $r_6$
Given,
$r_1=r_4<r_2<r_6$

Since, $r_1<r_2$, A is the catalyst for the reaction. $r_1=r_4$ implies that C does not affect the rate of reaction. but $r_6$ is highest so the catalyst A is more effective in presence of C. So, C is a catalytic promoter of catalyst A.

(C)
$R\to P$ rate = $r_1$
$R\stackrel{A}{\to }P$ rate = $r_2$
$R\stackrel{D}{\to }P$ rate = $r_5$
$R\stackrel{A+D}{\to }P$ rate = $r_7$

Given,
$r_1=r_5<r_7<r_2$
since $r_1=r_5$, D has no effect on the rate of reaction. $r_7<r_2$ implies that in presence of D the catalytic activity of A is decreased. So, D is catalytic poison of A

(D) A is a catalytic promotes of catalyst C
$R\to P$ i rate = $r_1$
$R\stackrel{A}{\to }P$ i rate = $r_2$
$R\stackrel{C}{\to }P$ i rate = $r_4$
$R\stackrel{A+C}{\to }P$ i rate = $r_6$
$r_1=r_4<r_2<r_6$
$r_1=r_4$ So, C does not affect the rate of reaction.
Hence, C is not a catalyst. $r_1<r_2$ implies that A is a catalyst. $r_2<r_6$ shows C is a catalytic promoter of catalyst A.

Hence, option (d) is the incorrect statement.