Consider the equilibrium:
$\sf CH_{4(g)}\ +\ H_2O_{(g)} = CO_{(g)}\ +\ 3H_{2(g)}$ $\sf K_{eq}=5.7$

A 1.0 L container is filled with 2.5 mol CH4 , 0.25 mol H2O, 1.2 mol CO and 2.2 mol H2.

Which of the following occurs?

Reaction Proceeds Pressure
a. left increases
b. left decreases
c. right increases
d. right decreases

Question

Consider the equilibrium:
$\sf CH_{4(g)}\ +\ H_2O_{(g)} = CO_{(g)}\ +\ 3H_{2(g)}$  $\sf K_{eq}=5.7$

 A 1.0 L container is filled with 2.5 mol CH4 , 0.25 mol H2O, 1.2 mol CO and 2.2 mol H2.

Which of the following occurs?


	Reaction Proceeds		Pressure
a.	left				increases
b.	left				decreases
c.	right				increases	
d. 	right				decreases

anonymous · Answer

First thing I did is, I calculated the $\sf K_{eq}$ using the given data and I got 20.44. Since this $\sf K_{eq}$ is greater than the given $\sf K_{eq}$, which 5.7, I assume that the equilibrium will shift to the Right and for that to happen, the Pressure should decrease. My answer is D but it is not right. I just need explanation why the right answer is A, why would it shift to LEFT??

anonymous · Answer

@aaronq @chmvijay @SolomonZelman @abb0t @Abhisar

abb0t · Answer

Keep in mind that when you find K$\sf _{eq}$, what does that equal? products over reactants.
That means that the there's more producte being formed than reactants.

abb0t · Answer

I think you mean that you calculated Q, not K.

anonymous · Answer

aah omg Thank YOU!! :D
Q is greater than K, so it means it should shift to the left in order to reach equilibrium, because the product-to-reactant ratio is high!! therefore, pressure should increase because it will favour the ones that has lesser gas moles...