At -20^{circ} C and 1 atm pressure,a cylinde | vedclass

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(C) For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = 2 - (1

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) For the reaction $H_{2(g)} + I_{2(g)} ightleftharpoons 2 HI_{(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = 2 - (1 + 1) = 0$. The equilibrium constant $K_p$ is related to the partial pressures of the reactants and products. Since $\Delta n_g = 0$,$K_p = K_c$. Given that the cylinder contains an equal number of moles of $H_2$,$I_2$,and $HI$,let $n_{H_2} = n_{I_2} = n_{HI} = n$. The partial pressure of any component $i$ is $P_i = \frac{n_i}{n_{total}} 	imes P_{total}$. $K_p = \frac{(P_{HI})^2}{P_{H_2} 	imes P_{I_2}} = \frac{(\frac{n}{3n} 	imes P_{total})^2}{(\frac{n}{3n} 	imes P_{total}) 	imes (\frac{n}{3n} 	imes P_{total})} = \frac{(\frac{1}{3} P_{total})^2}{(\frac{1}{3} P_{total}) 	imes (\frac{1}{3} P_{total})} = 1$. Given $K_p = x 	imes 10^{-1}$,we have $1 = x 	imes 10^{-1}$,which implies $x = 10$.

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