Select the incorrect statement.

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(NONE) For the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$,the rate expression is $-\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$

Vedclass · Accepted Answer

For $N_2 + 3H_2 \rightleftharpoons 2NH_3$; if rate of formation of $NH_3$ is $0.001 \ kg \ hr^{-1}$,then rate of consumption of $H_2$ is $0.0015 \ kg \ hr^{-1}$.

Vedclass · Answer

(NONE) For the reaction $N_2 + 3H_2 ightleftharpoons 2NH_3$,the rate expression is $-\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.Given rate of formation of $NH_3$ is $0.001 \ kg \ hr^{-1}$.To find the rate of consumption of $H_2$ in $kg \ hr^{-1}$,we use the stoichiometry: Rate of consumption of $H_2 = \frac{3}{2} 	imes (	ext{Rate of formation of } NH_3)$.Rate of consumption of $H_2 = \frac{3}{2} 	imes 0.001 \ kg \ hr^{-1} = 0.0015 \ kg \ hr^{-1}$.Since the units are consistent $(kg \ hr^{-1})$,the statement in option $C$ is correct.However,the rate of reaction always increases with temperature for both exothermic and endothermic reactions due to the increase in the number of molecules with energy greater than $E_a$.Upon re-evaluating the options,all statements $A, B, C,$ and $D$ are scientifically correct. There is no incorrect statement provided.

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