Find out ln K_{eq} for the formation of NO_2 | vedclass

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

Question

(A) The standard Gibbs free energy change for the reaction is calculated as:$\Delta G^o = \Delta G^o_f(NO_2) - [\Delta G^o_f(NO) + \frac{1}{2} \Delta

Vedclass · Accepted Answer

$\frac{35 	imes 10^3}{8.314 	imes 298}$

Vedclass · Answer

(A) The standard Gibbs free energy change for the reaction is calculated as:$\Delta G^o = \Delta G^o_f(NO_2) - [\Delta G^o_f(NO) + \frac{1}{2} \Delta G^o_f(O_2)]$Substituting the given values:$\Delta G^o = 52.0 - [87.0 + 0] = -35.0 \ kJ/mol = -35 	imes 10^3 \ J/mol$The relationship between standard Gibbs free energy and the equilibrium constant is:$\Delta G^o = -RT \ln K_{eq}$Rearranging for $\ln K_{eq}$:$\ln K_{eq} = -\frac{\Delta G^o}{RT}$Substituting the values:$\ln K_{eq} = -\frac{-35 	imes 10^3 \ J/mol}{8.314 \ J/mol \ K 	imes 298 \ K} = \frac{35 	imes 10^3}{8.314 	imes 298}$

Find out $\ln K_{eq}$ for the formation of $NO_2$ from $NO$ and $O_2$ at $298 \ K$.
$NO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{2(g)}$
Given:
$\Delta G^o_f (NO_2) = 52.0 \ kJ/mol$
$\Delta G^o_f (NO) = 87.0 \ kJ/mol$
$\Delta G^o_f (O_2) = 0 \ kJ/mol$

Similar Questions

Find the equilibrium constant for the reaction below at $298 \ K$. $2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$ given that $\Delta H^o = 18.4512 \ kcal$ and $\Delta S^o = 29.16 \ cal/K$.

At $300 \ K$,the equilibrium constant for a reaction is $10$. The standard free energy change (in $kJ \ mol^{-1}$) for the reaction is

The equilibrium constant for a reaction is $10$. What will be the value of $\Delta G^{\theta}$? $R = 8.314 \, J \, K^{-1} \, mol^{-1}, T = 300 \, K$

Consider the following reaction at $298 \ K$.
$\frac{3}{2} O_{2(g)} \rightleftharpoons O_{3(g)} ; K_{P} = 2.47 \times 10^{-29}$.
$\Delta_{r} G^{\ominus}$ for the reaction is $ . . . . . . \ kJ$. (Given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

$A$ reaction attains equilibrium when the free energy change accompanying it is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Find out $\ln K_{eq}$ for the formation of $NO_2$ from $NO$ and $O_2$ at $298 \ K$.$NO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{2(g)}$Given:$\Delta G^o_f (NO_2) = 52.0 \ kJ/mol$$\Delta G^o_f (NO) = 87.0 \ kJ/mol$$\Delta G^o_f (O_2) = 0 \ kJ/mol$