For the reaction N_2 + 3H_2 rightarrow 2NH_3 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The rate of reaction is given by the expression: $Rate = -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt}$ Given

Vedclass · Accepted Answer

$1.25 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$

Vedclass · Answer

(B) The rate of reaction is given by the expression: $Rate = -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt}$ Given that the rate of formation of $NH_3$ is $\frac{d[NH_3]}{dt} = 2.5 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$. Equating the rate of disappearance of $N_2$ to the rate of formation of $NH_3$: $-\frac{d[N_2]}{dt} = \frac{1}{2} 	imes \frac{d[NH_3]}{dt}$ $-\frac{d[N_2]}{dt} = \frac{1}{2} 	imes (2.5 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1})$ $-\frac{d[N_2]}{dt} = 1.25 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$

For the reaction $N_2 + 3H_2 \rightarrow 2NH_3$ in the Haber process,the rate of formation of $NH_3$ is $2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$. What is the rate of disappearance of $N_2$?

Similar Questions

Which of the following is the fastest reaction?

For the reaction $CH_3COOCH_3 + H_2O \xrightarrow{H^{+}} CH_3COOH + CH_3OH$,the progress of the reaction is followed by:

Ammonia and oxygen react at high temperature as in the reaction,$4 NH_{3(g)} + 5 O_{2(g)} \rightarrow 4 NO_{(g)} + 6 H_2O_{(g)}$. If the rate of formation of $NO$ is $3.6 \times 10^{-3} \ mol \ L^{-1} \ sec^{-1}$,calculate the rate of formation of water.

Consider the following reaction: $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$. The rate of this reaction in terms of $N_2$ at $T \ K$ is $\frac{-d[N_2]}{dt} = 0.02 \ mol \ L^{-1} \ s^{-1}$. What is the value of $\frac{-d[H_2]}{dt}$ (in units of $mol \ L^{-1} \ s^{-1}$) at the same temperature?

The term $\left( -\frac{dc}{dt} \right)$ in a rate equation refers to the:

Vedclass Test Series

Exam Paper Generator

Online Exam Module