For the reversible reaction N_{2(g)} + 3H_{2( | vedclass

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Question

(A) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n}$.Rearranging for $K_c$: $K_c = K_p / (RT)^{\Delta n}$.

Vedclass · Accepted Answer

$1.44 	imes 10^{-5} / (0.082 	imes 773)^{-2}$

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(A) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n}$.Rearranging for $K_c$: $K_c = K_p / (RT)^{\Delta n}$.For the reaction $N_{2(g)} + 3H_{2(g)} ightleftharpoons 2NH_{3(g)}$,the change in moles of gas is $\Delta n = 2 - (1 + 3) = -2$.The temperature $T = 500 \, ^\circ C = 500 + 273 = 773 \, K$.The gas constant $R = 0.082 \, 	ext{L atm K}^{-1} 	ext{mol}^{-1}$.Substituting the values: $K_c = 1.44 	imes 10^{-5} / (0.082 	imes 773)^{-2}$.

For the reversible reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ at $500 \, ^\circ C$,the value of $K_p$ is $1.44 \times 10^{-5}$. When the partial pressure is measured in atmospheres,the value of $K_c$ in $\text{mol L}^{-1}$ is:

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