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$$\begin{alignat}{2} \ce {CO2 (g) + 3H2 (g) \;& <=> CH3OH (g) + H2O (g)}\qquad&&{k_1=\;?} \\ \ce {CO (g) + H2O (g) \;& <=> CO2(g) + H2(g)}\qquad&&{k_2= 1.0\times10^5} \\ \ce {CO (g) + 2H2 (g) \;& <=> CH3OH(g)}\qquad&&{k_3= 1.4\times10^7} \end{alignat}$$

What is the value of $k_1$? I have tried reversing equation two and then multiplying $k_2$ and $k_3$ together, but I am not sure how to work this problem out.

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Remember that when you reverse the reaction the equilibrium constant changes. For the general gas-phase reaction $\ce{A(g) + B(g) <=> C(g) + D(g)}$ the equilibrium constant expression is $$K_\text{f} = \frac{p_\ce{C} p_\ce{D} }{p_\ce{A} p_\ce{B}}$$ (Strictly, it's activities, not partial pressures, but the principle is the same.) The reversed reaction $\ce{C(g) + D(g) <=> A(g) + B(g)}$ has the equilibrium constant expression $$K_\text{r} = \frac{p_\ce{A} p_\ce{B} }{p_\ce{C} p_\ce{D}} = \frac{1}{K_\text{f}}$$

You are correct in multiplying the equilibrium constants to get the equilibrium constant for the combined reaction.

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  • $\begingroup$ do I multiply k2 and k3 together to get kf? $\endgroup$ Oct 2 '15 at 1:49
  • $\begingroup$ I figured it out. The answer is 140. $\endgroup$ Oct 2 '15 at 17:46
  • $\begingroup$ @DaveLaRumbe That's right. $\endgroup$
    – j_foster
    Oct 3 '15 at 1:30

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