(D) The equilibrium expression for the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ is given by $K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}}$.Given

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(D) The equilibrium expression for the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ is given by $K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}}$.Given $K_p = 0.113 \ atm$ and $P_{N_2O_4} = 0.2 \ atm$.Substituting these values into the expression: $0.113 = \frac{(P_{NO_2})^2}{0.2}$.$(P_{NO_2})^2 = 0.113 \times 0.2 = 0.0226$.$P_{NO_2} = \sqrt{0.0226} \approx 0.15 \ atm$.

Class 11 Chemistry 1 Equilibrium: Chemical Equilibrium Kp and Kc Relationship

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At $298 \ K$,the value of $K_p$ for $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ is $0.113 \ atm$. The partial pressure of $N_2O_4$ at equilibrium is $0.2 \ atm$. What is the partial pressure (in $atm$) of $NO_2$ at equilibrium?

AP EAMCET 2025 All AP EAMCET

A
$0.05$
B
$0.075$
C
$0.30$
D
$0.15$

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1 Equilibrium: Chemical Equilibrium

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Kp and Kc Relationship

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At $298 \ K$,the value of $K_p$ for $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ is $0.113 \ atm$. The partial pressure of $N_2O_4$ at equilibrium is $0.2 \ atm$. What is the partial pressure (in $atm$) of $NO_2$ at equilibrium?

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Explain why the partial pressure $p$ of a gas is proportional to its concentration $c$.

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