CuO(s) + H\(_2\)(g) ↔ Cu(s) + H\(_2\)O(g)
What is the effect of increasing the pressure on the equilibrium reaction above?

a

The equilibrium is shifted to the left

b

The equilibrium is shifted to the right

c

There is no effect

d

More H2(g) is produced

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Explanation

Correct Option
c

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Discussions (17)

badformular
10 years ago

yes becos there is no diffrence in the number of moles on both sides

MosesCharles
4 years ago

Hey guys, the no. Of moles must not be the same on both side of the equation.

CHRIST IS KING
2 months ago

new change in pressure does not affect reaction that has equal volume on both sides

Since there is hydrogen gas in the reaction above there will be an effect of pressure in the reaction. Hence since the reverse reaction has a greater number of gaseous molecules, an increase in pressure will favour the forward reaction and push the equilibrium position to the right.
Thanks

Domnickado
3 years ago

Like seriously at this I'm not sure if I can trust MYSCHOOL anymore, the answer is definitely B.
The effect of pressure applys as long as one of the species is gaseous

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