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Question 13.7: Relating the Equilibrium Constants Kp and Kc Hydrogen is pro......

Relating the Equilibrium Constants K_p and K_c

Hydrogen is produced industrially by the steam–hydrocarbon re-forming process. The reaction that takes place in the first step of this process is

\mathrm{H}_2 \mathrm{O}(g)+\mathrm{CH}_4(g) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_2(g)

(a) If K_c = 3.8 × 10^{-3} at 1000 K, what is the value of K_p at the same temperature?

(b) If K_p = 6.1 × 10^4 at 1125 °C, what is the value of K_c at 1125 °C?

STRATEGY

To calculate K_p from K_c, or vice versa, use the equation K_p = K_c(RT)^{Δn} , where R must be in units of (L . atm)/(K . mol), T is the temperature in kelvin, and Δn is the number of moles of gaseous products minus the number of moles of gaseous reactants.

worked example 13.7
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(a) For this reaction, Δn = (1 + 3) – (1 + 1) = 2. Therefore,

K_{\mathrm{p}}=K_{\mathrm{c}}(R T)^{\Delta n}=K_{\mathrm{c}}(R T)^2=\left(3.8 \times 10^{-3}\right)[(0.08206)(1000)]^2=26

(b) Solving the equation K_p = K_c(RT)^2 for K_c gives

K_{\mathrm{c}}=\frac{K_{\mathrm{p}}}{(R T)^2}=\frac{6.1 \times 10^4}{[(0.08206)(1398)]^2}=4.6

Note that the temperature in these equations is the absolute temperature; 1125 °C corresponds to 1125 + 273 = 1398 K.

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