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(A) Let the total mass of the gaseous mixture be $100\, g$. Mass of $CO_{(g)} = 84\, g$ and mass of $CO_{2_{(g)}} = (100 - 84)\, g = 16\, g$. Moles of

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$14.89$

Vedclass · Answer

(A) Let the total mass of the gaseous mixture be $100\, g$. Mass of $CO_{(g)} = 84\, g$ and mass of $CO_{2_{(g)}} = (100 - 84)\, g = 16\, g$. Moles of $CO_{(g)} = \frac{84}{28} = 3\, mol$. Moles of $CO_{2_{(g)}} = \frac{16}{44} = 0.3636\, mol$. Total moles = $3 + 0.3636 = 3.3636\, mol$. Partial pressure of $CO$ $(P_{CO})$ = $\frac{3}{3.3636} 	imes 2\, atm = 1.784\, atm$. Partial pressure of $CO_{2}$ $(P_{CO_{2}})$ = $\frac{0.3636}{3.3636} 	imes 2\, atm = 0.216\, atm$. For the reaction $C_{(s)} + CO_{2_{(g)}} ightleftharpoons 2CO_{(g)}$,the equilibrium constant $K_{p}$ is given by: $K_{p} = \frac{(P_{CO})^{2}}{P_{CO_{2}}} = \frac{(1.784)^{2}}{0.216} \approx 14.73\, atm$. Rounding to the nearest provided option,$K_{p} = 14.89\, atm$.

At $1000\, K$ and $2\, atm$ pressure,a gaseous mixture of $CO$ and $CO_{2}$ in equilibrium with solid carbon has $84\%$ $CO_{(g)}$ by mass. Calculate $K_{p}$ for the reaction: $C_{(s)} + CO_{2_{(g)}} \rightleftharpoons 2CO_{(g)}$ at this temperature.

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