The reaction below has an equilibrium constant of
Kp=2.26×104 at 298 K.
CO(g)+2H2(g)⇌CH3OH(g)
A) Calculate Kp for the reaction below.
12CH3OH(g)⇌12CO(g)+H2(g)
B) Calculate Kp for the reaction below.
2CO(g)+4H2(g)⇌2CH3OH(g)
C) Calculate Kp for the reaction below.
3CH3OH(g)⇌3CO(g)+6H2(g)
A)
reaction givev is not balanced
It will be
12CH3OH(g)⇌12CO(g)+24H2(g)
It can be obtained from original given reaction by reversing and then multiplying it by 12
Kp’ = (1/kp)^12
= (1/(2.26*10^4))^12
=5.63*10^-53
B)
It can be obtained from original given reaction by multiplying it by 2
Kp’ = (kp)^2
= (2.26*10^4)^2
=5.11*10^8
C)
It can be obtained from original given reaction by reversing and then multiplying it by 3
Kp’ = (1/kp)^3
= (1/(2.26*10^4))^3
=8.66*10^-14
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