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?

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.

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.