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(C) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n_g}$.Rearranging this,we get $\frac{K_c}{K_p} = (RT)^{-\

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$400$

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(C) The relationship between $K_p$ and $K_c$ is given by the formula: $K_p = K_c(RT)^{\Delta n_g}$.Rearranging this,we get $\frac{K_c}{K_p} = (RT)^{-\Delta n_g}$.For the reaction $N_{2(g)} + 3H_{2(g)} ightleftharpoons 2NH_{3(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = n_p - n_r = 2 - (1 + 3) = 2 - 4 = -2$.Substituting the values into the equation: $\frac{K_c}{K_p} = (RT)^{-(-2)} = (RT)^2$.Given $\frac{K_c}{K_p} = 1076$,we have $(RT)^2 = 1076$.Taking the square root of both sides: $RT = \sqrt{1076} \approx 32.8$.Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$,we have $0.082 	imes T = 32.8$.$T = \frac{32.8}{0.082} = 400 \ K$.

For the following given equilibrium reaction,$\frac{K_{c}}{K_{p}}$ is equal to $1076$ at $T \ K$. What is the value of $T$ (in $K$)? $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

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For the following given equilibrium reaction,$\frac{K_{c}}{K_{p}}$ is equal to $1076$ at $T \ K$. What is the value of $T$ (in $K$)? $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$