In a reaction N_{2(g)} + 3H_{2(g)} to 2NH_{3( | vedclass

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

Question

(C) For the reaction $N_{2(g)} + 3H_{2(g)} 	o 2NH_{3(g)}$,the rate expression is given by:Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt}

Vedclass · Accepted Answer

$1.25 	imes 10^{-4}, 3.75 	imes 10^{-4}$

Vedclass · Answer

(C) For the reaction $N_{2(g)} + 3H_{2(g)} 	o 2NH_{3(g)}$,the rate expression is given by:Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$Given,rate of appearance of $NH_3$ is $\frac{d[NH_3]}{dt} = 2.5 	imes 10^{-4} \, mol \, L^{-1} \, s^{-1}$.Rate of reaction $= \frac{1}{2} 	imes \frac{d[NH_3]}{dt} = \frac{1}{2} 	imes 2.5 	imes 10^{-4} = 1.25 	imes 10^{-4} \, mol \, L^{-1} \, s^{-1}$.Rate of disappearance of $H_2$ is $-\frac{d[H_2]}{dt}$.From the rate expression,$-\frac{d[H_2]}{dt} = \frac{3}{2} \frac{d[NH_3]}{dt} = \frac{3}{2} 	imes 2.5 	imes 10^{-4} = 3.75 	imes 10^{-4} \, mol \, L^{-1} \, s^{-1}$.

In a reaction $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$,the rate of appearance of $NH_3$ is $2.5 \times 10^{-4} \, mol \, L^{-1} \, s^{-1}$. The rate of reaction and rate of disappearance of $H_2$ will be (in $mol \, L^{-1} \, s^{-1}$):

Similar Questions

In the circuit shown below,the ammeter reading is zero. Then the value of the resistance $R$ is (in $Omega$)

$100 \ mL$ of $0.6 \ M$ acetic acid is shaken with $2 \ g$ of activated charcoal. After adsorption,the final concentration of the solution becomes $0.5 \ M$. The amount of acetic acid adsorbed per gram of carbon is .......... $g$.

Two concentric coils of $10$ turns each are placed in the same plane. Their radii are $20 \ cm$ and $40 \ cm$ and carry $0.2 \ A$ and $0.3 \ A$ current respectively in opposite directions. The magnetic induction (in $T$) at the centre is

$A$ polygon has $54$ diagonals. Then,the number of its sides is

Vedclass Test Series

Exam Paper Generator

Online Exam Module