For a reversible reaction $A \rightleftharpoons B$,the pre-exponential factor is the same for both the forward and backward reactions and has a value of $20 \ s^{-1}$. If the enthalpy change for the forward reaction is $-41.5 \ kJ \ mol^{-1}$,the value of the equilibrium constant at $500 \ K$ is:

  • A
    $e^{10}$
  • B
    $e^9$
  • C
    $e^8$
  • D
    $e^7$

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At a certain temperature,only $50\%$ $HI$ is dissociated into $H_2$ and $I_2$ at equilibrium. The equilibrium constant is:

For the reactions:
$2NO + O_2 \rightleftharpoons 2NO_2$; $K_1$
$4NO + 2Cl_2 \rightleftharpoons 4NOCl$; $K_2$
$NO_2 + \frac{1}{2}Cl_2 \rightleftharpoons NOCl + \frac{1}{2}O_2$; $K_3$
Where $K_1, K_2, K_3$ are equilibrium constants,then $K_3^2$ is equal to:

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For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,$K_p = 0.492 \ atm$ at $300 \ K$. $K_c$ for the reaction at same temperature is . . . . . . $\times 10^{-2}$. (Given: $R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$)

At $400 \ K$,in a $1.0 \ L$ vessel,$N_2O_4$ is allowed to attain equilibrium,$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. At equilibrium,the total pressure is $600 \ mm \ Hg$,when $20 \%$ of $N_2O_4$ is dissociated. The value of $K_p$ for the reaction is

Consider the partial decomposition of $A$ as:
$2A_{(g)} \rightleftharpoons 2B_{(g)} + C_{(g)}$
At equilibrium,a $700 \ mL$ gaseous mixture contains $100 \ mL$ of gas $C$ at $10 \ atm$ and $300 \ K$. What is the value of $K_P$ for the reaction?

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