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The reaction below has an equilibrium constant of Kp=2
The reaction below has an equilibrium constant of
Kp=2.26×104 at 298 K.
CO(g)+2H2(g)?CH3OH(g)
Part A: Calculate Kp for the reaction below.
1/2CH3OH(g)?1/2CO(g)+H2(g)
Part B: Predict whether reactants or products will be favored at equilibrium in the reaction above.
Part C: Calculate Kp for the reaction below.
2CO(g)+4H2(g)?2CH3OH(g)
Part D: Predict whether reactants or products will be favored at equilibrium in the reaction above.
Part E: Calculate Kp for the reaction below.
2CH3OH(g)?2CO(g)+4H2(g)
Part F: Predict whether reactants or products will be favored at equilibrium in the reaction above.
Expert Solution
A)
This reaction is -1/2 * given reaction
So,
Kp’ = (Kp)^-1/2
= (2.26*10^4)^-1/2
= (1/(2.26*10^4))^1/2
= (4.42*10^-5)^1/2
= 6.65*10^-3
Answer: 6.65*10^-3
B)
Since Kp is less than 1, reactants will be favoured
C)
This reaction is -2 * given reaction
So,
Kp’ = (Kp)^2
= (2.26*10^4)^2
= 5.11*10^8
Answer: 5.11*10^8
D)
Since Kp is greater than 1, products will be favoured
E)
This reaction is -2 * given reaction
So,
Kp’ = (Kp)^-2
= (2.26*10^4)^-2
= (1/(2.26*10^4))^2
= (4.42*10^-5)^2
= 1.95*10^-9
Answer: 1.95*10^-9
F)
Since Kp is less than 1, reactants will be favoured
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