Calculate the standard Gibbs energy change $(\Delta G^\circ)$ for a gaseous reaction at $298 \text{ K}$ if the equilibrium constant $K_p$ is $3.5 \times 10^{17}$ and the gas constant $R$ is $8.314 \text{ J K}^{-1} \text{mol}^{-1}$.

  • A
    -$100.1 \text{ kJ mol}^{-1}$
  • B
    -$79.5 \text{ kJ mol}^{-1}$
  • C
    -$71.4 \text{ kJ mol}^{-1}$
  • D
    -$89.5 \text{ kJ mol}^{-1}$

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Similar Questions

For the reaction $A_{(g)} \rightarrow B_{(g)},$ the value of the equilibrium constant at $300 \ K$ and $1 \ atm$ is equal to $100.0.$ The value of $\Delta_{r}G^{\circ}$ for the reaction at $300 \ K$ and $1 \ atm$ in $J \ mol^{-1}$ is $-xR,$ where $x$ is ........... (Rounded off to the nearest integer) ($R = 8.31 \ J \ mol^{-1} K^{-1}$ and $\ln 10 = 2.3$)

For the reaction,$2 NH_{3(g)} + CO_{2(g)} \rightleftharpoons NH_2CONH_{2(aq)} + H_2O_{(l)}$,find the value of the equilibrium constant at $295 \ K$. Given,the standard Gibbs energy change at the given temperature is $13.9 \ kJ \ mol^{-1}$.

The $INCORRECT$ match in the following is

For the equilibrium $H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$ at $1 \ atm$ and $298 \ K$,which of the following statements is correct?

At $298 \ K$, if the standard Gibbs energy change $\Delta_r G^{\ominus}$ of a reaction is $-115 \ kJ \ mol^{-1}$, the value of $\log_{10} K_{p}$ will be $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$.

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