For the reversible reaction in equilibrium:
$N_{2(g)} + O_{2(g)} \underset{k_2}{\overset{k_1}{\longleftrightarrow}} 2NO_{(g)}$
Given $C_0 = C e^{-2.1 \times 10^{-3}t}$ for the forward reaction and $C'_0 = C' e^{-4.2 \times 10^{-4}t}$ for the backward reaction,calculate the equilibrium constant $K_c$ for the above reaction.
$N_{2(g)} + O_{2(g)} \underset{k_2}{\overset{k_1}{\longleftrightarrow}} 2NO_{(g)}$
Given $C_0 = C e^{-2.1 \times 10^{-3}t}$ for the forward reaction and $C'_0 = C' e^{-4.2 \times 10^{-4}t}$ for the backward reaction,calculate the equilibrium constant $K_c$ for the above reaction.
- A$5.0$
- B$2.0$
- C$0.5$
- D$0.2$
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At $450 \, K$,$K_{p} = 2.0 \times 10^{10} \, bar^{-1}$ for the given reaction at equilibrium.
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What is $K_{c}$ at this temperature?
$2 \, SO_{2(g)} + O_{2(g)} \longleftrightarrow 2 \, SO_{3(g)}$
What is $K_{c}$ at this temperature?
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View SolutionFor the following three equilibria $X, Y$,and $Z$ in the gaseous state at $300 \ K$,the increasing order of the ratio of $K_p$ to $K_c$ is:
$X: 2SO_2 + O_2 \rightleftharpoons 2SO_3$
$Y: PCl_5 \rightleftharpoons PCl_3 + Cl_2$
$Z: 2HI \rightleftharpoons H_2 + I_2$
$X: 2SO_2 + O_2 \rightleftharpoons 2SO_3$
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