$(i)$ $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$ at $298 \ K$ has $\Delta G^{\Theta} = -16.5 \ kJ \ mol^{-1}$. Find $K_p$.
$(ii)$ At $298 \ K$,for $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,calculate $K_p$ and $\Delta G^{\Theta}$.
$(ii)$ At $298 \ K$,for $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,calculate $K_p$ and $\Delta G^{\Theta}$.
(N/A) $(i)$ Using the relation $\Delta G^{\Theta} = -RT \ln K_p$:
$-16500 \ J \ mol^{-1} = -(8.314 \ J \ K^{-1} \ mol^{-1}) \times (298 \ K) \times \ln K_p$
$\ln K_p = \frac{16500}{8.314 \times 298} \approx 6.658$
$K_p = e^{6.658} \approx 779.4$
$(ii)$ The reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ is twice the first reaction.
Therefore,$\Delta G^{\Theta}_{new} = 2 \times (-16.5 \ kJ \ mol^{-1}) = -33 \ kJ \ mol^{-1}$.
$K_{p, new} = (K_p)^2 = (779.4)^2 \approx 6.07 \times 10^5$.
$-16500 \ J \ mol^{-1} = -(8.314 \ J \ K^{-1} \ mol^{-1}) \times (298 \ K) \times \ln K_p$
$\ln K_p = \frac{16500}{8.314 \times 298} \approx 6.658$
$K_p = e^{6.658} \approx 779.4$
$(ii)$ The reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ is twice the first reaction.
Therefore,$\Delta G^{\Theta}_{new} = 2 \times (-16.5 \ kJ \ mol^{-1}) = -33 \ kJ \ mol^{-1}$.
$K_{p, new} = (K_p)^2 = (779.4)^2 \approx 6.07 \times 10^5$.
Explore More
Similar Questions
At $1000 \ K$,the equilibrium constant $K_C$ for the reaction $2 \ NOCl_{(g)} \rightleftharpoons 2 \ NO_{(g)} + Cl_{2(g)}$ is $4.0 \times 10^{-6} \ mol \ L^{-1}$. The $K_P$ (in bar) at the same temperature is $\left(R=0.083 \ L \ bar \ K^{-1} \ mol^{-1}\right)$
For the reaction: $HI_{(g)} \rightleftharpoons \frac{1}{2}H_{2(g)} + \frac{1}{2}I_{2(g)}$,the equilibrium constant is $8$. What is the equilibrium constant for the reaction: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$?
Medium
View SolutionThe value of $\log \ K$ for the reaction $A \rightleftharpoons B$ at $298 \ K$ is (Nearest integer).
Given: $\Delta H^{\circ} = -54.07 \ kJ \ mol^{-1}$
$\Delta S^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$
(Take $2.303 \times 8.314 \times 298 = 5705$)
Given: $\Delta H^{\circ} = -54.07 \ kJ \ mol^{-1}$
$\Delta S^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$
(Take $2.303 \times 8.314 \times 298 = 5705$)
The standard equilibrium constant,$K_p$ at $298 \, K$ for the reaction,$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ is $5.8 \times 10^5$. The value of standard equilibrium constant,if the concentration of gases is expressed in terms of $mol/L$,will be:
[Given : $R = 0.08314 \, L \, bar \, K^{-1} \, mol^{-1}$]
[Given : $R = 0.08314 \, L \, bar \, K^{-1} \, mol^{-1}$]
MediumNEET 2017
View SolutionFor the reaction $A_{(g)} \rightleftharpoons 2 B_{(g)}$,the backward reaction rate constant is higher than the forward reaction rate constant by a factor of $2500$,at $1000 \ K$. [Given : $R = 0.0831 \ L \ atm \ mol^{-1} \ K^{-1}$] $K_p$ for the reaction at $1000 \ K$ is
DifficultNEET 2025
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo