One mole of $N_2O_4(g)$ is taken in a closed container at $1 \ atm$ and $300 \ K$. When it is heated to $600 \ K$,$20 \%$ of $N_2O_4(g)$ dissociates into $NO_2(g)$. The resulting pressure is .......... $atm$.

  • A
    $1.2$
  • B
    $2.4$
  • C
    $2$
  • D
    $1$

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Similar Questions

$6.9 \ g$ of $N_2O_4$ is taken in a $0.5 \ L$ closed vessel at $400 \ K$. For the equilibrium $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the total pressure at equilibrium is $9.15 \ atm$. Calculate $K_c$,$K_p$,and the partial pressure of each component.

Explain the relationship between the time taken to reach equilibrium and thermodynamics.

The equilibrium constant of a reaction at $298 \ K$ is $5 \times 10^{-3}$ and at $1000 \ K$ is $2 \times 10^{-5}$. What is the sign of $\Delta H$ for the reaction?

In a closed vessel,the decomposition of solid $Mg(HCO_3)_2$ is given as follows:
$Mg(HCO_3)_{2(s)} \rightleftharpoons MgCO_{3(s)} + CO_{2(g)} + H_2O_{(g)}$
Given $K_p = 64 \ bar^2$ and the mole fraction of $CO_2$ is $0.8$,calculate the total pressure at equilibrium. (in $bar$)

At $298 \ K$,
$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g), K_1 = 4 \times 10^5$
$N_2(g) + O_2(g) \rightleftharpoons 2NO(g), K_2 = 1.6 \times 10^{12}$
$H_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O(g), K_3 = 1.0 \times 10^{-13}$
Based on the above equilibria,the equilibrium constant of the reaction,
$2NH_3(g) + \frac{5}{2}O_2(g) \rightleftharpoons 2NO(g) + 3H_2O(g)$
is $.......... \times 10^{-33}$ (Nearest integer).

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