Chapter 13

Q. 13.7

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?


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


Verified Solution

(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.