For the reaction at $25\,^{\circ}C$,$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,if $\Delta G^{\circ}_f$ for $N_2O_4$ and $NO_2$ are $23.49 \, kcal$ and $12.39 \, kcal$ respectively,then $K_p$ for the reaction is:
- A$113332$
- B$11.33$
- C$1.133$
- D$0.113$
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Similar Questions
Consider the following reversible chemical reactions:
$A_{2(g)} + B_{2(g)} \overset {K_1} \leftrightarrows 2AB_{(g)} ......(1)$
$6AB_{(g)} \overset {K_2} \leftrightarrows 3A_{2(g)} + 3B_{2(g)} ......(2)$
The relation between $K_1$ and $K_2$ is:
$A_{2(g)} + B_{2(g)} \overset {K_1} \leftrightarrows 2AB_{(g)} ......(1)$
$6AB_{(g)} \overset {K_2} \leftrightarrows 3A_{2(g)} + 3B_{2(g)} ......(2)$
The relation between $K_1$ and $K_2$ is:
DifficultJEE MAIN 2019
View SolutionThe value of $K_P / K_C$ for the reaction at $T(K)$ is:
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons CO_{2(g)}$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons CO_{2(g)}$
One mole of $PCl_{5(g)}$ was heated in a $1 \ L$ closed flask at $500 \ K$. At equilibrium,$0.1 \ mole$ of $Cl_{2(g)}$ was formed. What is its $K_{p}$ (in $atm$)? (Given $R=0.082 \ L \ atm \ mol^{-1} \ K^{-1}$)
In the gas phase reaction,$C_2H_4 + H_2 \rightleftharpoons C_2H_6$,the equilibrium constant can be expressed in units of
MediumAIPMT 1992
View SolutionThe equilibrium constant $(K_P)$ for the reaction $2SO_2 + O_2 \rightleftharpoons 2SO_3$ at $1000 \ K$ is $3.5$. The partial pressure of oxygen gas to give equal mole of $SO_2$ and $SO_3$ is $... \ atm$.
Medium
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