For the reaction {N_2}_{(g)} + 3{H_2}_{(g)} r | vedclass

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(B) The relationship between ${K_P}$ and ${K_C}$ is given by the formula: ${K_P} = {K_C}(RT)^{\Delta n}$.For the reaction ${N_2}_{(g)} + 3{H_2}_{(g)}

Vedclass · Accepted Answer

$3.56 	imes 10^8$

Vedclass · Answer

(B) The relationship between ${K_P}$ and ${K_C}$ is given by the formula: ${K_P} = {K_C}(RT)^{\Delta n}$.For the reaction ${N_2}_{(g)} + 3{H_2}_{(g)} ightleftharpoons 2{NH_3}_{(g)}$,the change in the number of moles of gas is $\Delta n = 2 - (1 + 3) = -2$.Given: ${K_P} = 5.8 	imes 10^5$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,and $T = 298 \ K$.Substituting the values: $5.8 	imes 10^5 = {K_C} 	imes (0.0821 	imes 298)^{-2}$.${K_C} = 5.8 	imes 10^5 	imes (0.0821 	imes 298)^2$.${K_C} = 5.8 	imes 10^5 	imes (24.4658)^2$.${K_C} = 5.8 	imes 10^5 	imes 598.57$.${K_C} \approx 3.47 	imes 10^8 \ mol^{-2} \ L^2$ (which is approximately $3.56 	imes 10^8$ depending on the value of $R$ used).

For the reaction ${N_2}_{(g)} + 3{H_2}_{(g)} \rightleftharpoons 2{NH_3}_{(g)}$,the equilibrium constant ${K_P}$ is $5.8 \times 10^5$ at $298 \ K$. Calculate the value of the equilibrium constant ${K_C}$ (in $mol^{-2} \ L^2$) at the same temperature.

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