For the reaction $2 H_{2(g)} + 2 NO_{(g)} \rightarrow N_{2(g)} + 2 H_2O_{(g)}$,the observed rate expression is $rate = k_f [NO]^2 [H_2]$. The rate expression of the reverse reaction is:

  • A
    $k_b [N_2] [H_2O]^2 / [NO]$
  • B
    $k_b [N_2] [H_2O]$
  • C
    $k_b [N_2] [H_2O]^2$
  • D
    $k_b [N_2] [H_2O]^2 / [H_2]$

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How is the direction of a reaction predicted?

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At $T(K)$,$K_c$ for the reaction $A_{2(g)} \rightleftharpoons B_{2(g)}$ is $99.0$. Two moles of $A_{2(g)}$ were heated to $T(K)$ in a $1 \ L$ closed flask to reach the above equilibrium. What are the concentrations (in $mol \ L^{-1}$) of $A_{2(g)}$ and $B_{2(g)}$ respectively at equilibrium?

$4.5$ moles each of hydrogen and iodine are heated in a sealed $10 \ L$ vessel. At equilibrium,$3$ moles of $HI$ are found. The equilibrium constant for ${H_2}_{(g)} + {I_2}_{(g)} \rightleftharpoons 2HI_{(g)}$ is

For the reaction equilibrium $N_2O_4(g) \rightleftharpoons 2NO_2(g)$,the concentrations of $N_2O_4$ and $NO_2$ at equilibrium are $4.8 \times 10^{-2} \ mol \ L^{-1}$ and $1.2 \times 10^{-2} \ mol \ L^{-1}$ respectively. The value of $K_c$ for the reaction is:

The following equilibrium constants are given:
$N_{2} + 3 H_{2} \rightleftharpoons 2 NH_{3} ; K_{1}$
$N_{2} + O_{2} \rightleftharpoons 2 NO ; K_{2}$
$H_{2} + \frac{1}{2} O_{2} \rightleftharpoons H_{2} O ; K_{3}$
The equilibrium constant for the oxidation of $2 \text{ mole}$ of $NH_{3}$ to give $NO$ is

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