$CoO_{2(g)} + H_{2(g)} \rightleftharpoons CoO_{(s)} + H_2O_{(g)} \,;\, K_1 = 67$
$CoO_{2(g)} + CO_{(g)} \rightleftharpoons CoO_{(s)} + CO_{2(g)} \,;\, K_2 = 490$
Then the equilibrium constant for the following reaction is ....
$CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$

  • A
    $0.137$
  • B
    $423$
  • C
    $7.31$
  • D
    $557$

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For the reversible reaction in equilibrium
$N_{2(g)} + O_{2(g)} \underset{k_2}{\overset{k_1}{\longleftrightarrow}} 2NO_{(g)}$
If the rate constant for the forward reaction is $k_1 = 2.1 \times 10^{-3} \ s^{-1}$ and for the backward reaction is $k_2 = 4.2 \times 10^{-4} \ s^{-1}$,then the equilibrium constant $K_c$ for the above reaction is:

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