The reaction between N_2 and H_2 to form ammo | vedclass

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(A) The balanced chemical equation is: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.The change in the number of moles of gaseous products and reacta

Vedclass · Accepted Answer

$1.5 	imes 10^{-5}$

Vedclass · Answer

(A) The balanced chemical equation is: $N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g)$.The change in the number of moles of gaseous products and reactants is $\Delta n = 2 - (1 + 3) = -2$.The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n}$.Given $K_c = 6 	imes 10^{-2}$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,and $T = 500 + 273 = 773 \ K$.Substituting the values: $K_p = 6 	imes 10^{-2} 	imes (0.0821 	imes 773)^{-2}$.$K_p = \frac{6 	imes 10^{-2}}{(63.4633)^2} \approx \frac{0.06}{4027.59} \approx 1.49 	imes 10^{-5} \approx 1.5 	imes 10^{-5}$.

The reaction between $N_2$ and $H_2$ to form ammonia has $K_c = 6 \times 10^{-2}$ at the temperature $500 \ ^oC$. The numerical value of $K_p$ for this reaction is

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