(C) For the reaction $N_{2}O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the change in the number of moles of gaseous species is $\Delta n_{g} = 2 - 1 = 1$.

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(C) For the reaction $N_{2}O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the change in the number of moles of gaseous species is $\Delta n_{g} = 2 - 1 = 1$.The relationship between $K_{P}$ and $K_{C}$ is given by $K_{P} = K_{C} \cdot (RT)^{\Delta n_{g}}$.Substituting the given values: $600.1 = 20.4 \times (0.0831 \times T)^{1}$.Solving for $T$: $T = \frac{600.1}{20.4 \times 0.0831} = \frac{600.1}{1.69524} \approx 353.99 \, K$.Rounding off to the nearest integer,we get $T = 354 \, K$.

Class 11 Chemistry 1 Equilibrium: Chemical Equilibrium Kp and Kc Relationship

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Consider the reaction $N_{2}O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. The temperature at which $K_{C}=20.4$ and $K_{P}=600.1$ is ....... $K$. (Round off to the nearest integer).
[Assume all gases are ideal and $R=0.0831 \, L \, bar \, K^{-1} \, mol^{-1}$]

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Consider the reaction $N_{2}O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. The temperature at which $K_{C}=20.4$ and $K_{P}=600.1$ is ....... $K$. (Round off to the nearest integer).[Assume all gases are ideal and $R=0.0831 \, L \, bar \, K^{-1} \, mol^{-1}$]

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One mole of $PCl_5$ is heated in a closed vessel of $1 \ L$ capacity. At equilibrium,$20\%$ of $PCl_5$ remains undissociated. Find the value of $K_c$.

For the reaction $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$ at $250 \ ^\circ C$,the value of $K_c$ is $26$. The value of $K_p$ at the same temperature will be:

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Consider the reaction $N_{2}O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. The temperature at which $K_{C}=20.4$ and $K_{P}=600.1$ is ....... $K$. (Round off to the nearest integer).
[Assume all gases are ideal and $R=0.0831 \, L \, bar \, K^{-1} \, mol^{-1}$]