Final answer:
K'p for the reverse reaction CH3OH(g) ↔ CO(g) + 2H2(g) is calculated by taking the reciprocal of Kp, resulting in 4.42×10^-5. For the reaction with altered stoichiometry to one twelfth, K'p is determined by taking the twelfth root of Kp, yielding a K'p value of approximately 3.08.
Step-by-step explanation:
To calculate K'p for the reverse reaction CH3OH(g) ⇌ CO(g) + 2H2(g), you simply take the reciprocal of the given equilibrium constant Kp. Since Kp = 2.26×104, the reversed reaction's K'p will be 1 / (2.26×104), which gives us a K'p value of approximately 4.42×10-5 to three significant figures.
For the reaction that involves one twelfth of the mole ratios (\(\frac{1}{2}\)CO(g) + \(\frac{1}{2}\)H2(g) ⇌ \(\frac{1}{2}\)CH3OH(g)), we need to take the twelfth root of the original Kp to find the new K'p, because the equilibrium constant expression depends on the stoichiometry of the reaction as expressed in the balanced equation. The twelfth root of 2.26×104 gives us K'p' ≈ 3.08, rounded to three significant figures.