One mole H_2O_{(g)} and one mole CO_{(g)} are | vedclass

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(B) For the reaction: $H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2_{(g)}} + CO_{2_{(g)}}$
$\Delta n_g = (1 + 1) - (1 + 1) = 0$
Since $\Delta n_g =

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$0.444$

Vedclass · Answer

(B) For the reaction: $H_2O_{(g)} + CO_{(g)} ightleftharpoons H_{2_{(g)}} + CO_{2_{(g)}}$
$\Delta n_g = (1 + 1) - (1 + 1) = 0$
Since $\Delta n_g = 0$,$K_p = K_c(RT)^{\Delta n_g} = K_c$.
Given that $40 \%$ of water reacted,the degree of reaction $\alpha = 0.4$.

	
		
			Species
			Initial moles
			Equilibrium moles
		
		
			$H_2O$
			$1.0$
			$0.6$
		
		
			$CO$
			$1.0$
			$0.6$
		
		
			$H_2$
			$0.0$
			$0.4$
		
		
			$CO_2$
			$0.0$
			$0.4$
		
	


Concentrations in $1 \ L$ flask: $[H_2O] = 0.6 \ M, [CO] = 0.6 \ M, [H_2] = 0.4 \ M, [CO_2] = 0.4 \ M$.
$K_c = \frac{[H_2][CO_2]}{[H_2O][CO]} = \frac{0.4 	imes 0.4}{0.6 	imes 0.6} = \frac{0.16}{0.36} = 0.444$.
Therefore,$K_p = 0.444$.

One mole $H_2O_{(g)}$ and one mole $CO_{(g)}$ are taken in a $1 \ L$ flask and heated to $725 \ K$. At equilibrium,$40 \%$ (by mass) of water reacted with $CO_{(g)}$ as follows: $H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2_{(g)}} + CO_{2_{(g)}}$. The value of $K_p$ is:

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