For the reaction taking place at a certain temperature $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$. If the equilibrium pressure is $3X \ bar$,then $\Delta _r G^o$ would be:

  • A
    $- RT \ln 9 - 3 RT \ln X$
  • B
    $RT \ln 4 - 3 RT \ln X$
  • C
    $- 3 RT \ln X$
  • D
    $- RT \ln 4 - 3 RT \ln X$

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Similar Questions

For the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$,the equilibrium constant at temperature $T$ is $4 \times 10^{-4}$. Find the value of $K_c$ for the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)}$.

$2N_2O_{4(g)} \rightleftharpoons 4NO_{2(g)}$ has $K_p = 0.15 \, atm$ at $298 \, K$. Calculate $K_p$ in $torr$ and $K_c$ in $mol \, L^{-1}$. ($1 \, atm = 760 \, torr$; $R = 0.082 \, L \, atm \, mol^{-1} \, K^{-1}$).

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$NH_4COONH_{2_{(s)}} \rightleftharpoons 2NH_{3_{(g)}} + CO_{2_{(g)}}$. If the equilibrium pressure is $3 \, atm$ for the above reaction,the $K_p$ for the reaction is:

One mole of $N_2O_4$ in a $1 \ L$ flask decomposes to attain the equilibrium $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. At the equilibrium the mole fraction of $NO_2$ is $1/2$. Hence $K_C$ will be:

The equilibrium constant for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ at temperature $T$ is $4 \times 10^{-4}$. The value of $K_c$ for the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{1}{2}O_{2(g)}$ at the same temperature is

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