Chemical EquilibriumHard
Question
Methanol (CH3OH) can be prepared from CO and H2 as CO(g) + 2H2(g) $\rightleftharpoons$CH3OH(g); KP = 6.23 × 10−3 at 500 K. What total pressure is required to convert 25% of CO to CH3OH at 500 K, if CO and H2 comes from the following reaction? CH4(g) + H2O(g) → CO(g) + 3H2(g)
Options
A.20.48 bar
B.21 bar
C.10.24 bar
D.5.12 bar
Solution
CO and H2 are initially in 1: 3 mole ratio, as they are formed by 2nd reaction.
CO + 2H2 $\rightleftharpoons$ CH3OH
Initial moles 1 3 0
Equilibrium moles 1 – 0.25 3 – 0.25 × 2 0.25
= 0.75 = 2.5
Total moles = 0.75 + 2.5 + 0.25 = 3.5
Now, $K_{P} = \frac{0.25}{0.75 \times (2.5)^{2}} \times \left( \frac{P}{3.5} \right)^{- 2} = 6.23 \times 10^{- 3}$
∴ P = 10.24 bar
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