Question

Consider the following generic reaction for which K = 0.00551:
2 A + 3D ⇄ E
Which of the following correctly applies to this reaction:

Products are favored at equilibrium.

Rate_forward = Rate_reverse

Keq =	[E]
	[A]2[D]3

Keq =	[A]2[D]3
	[E]

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Consider the following generic reaction for which K = 0.00551:
2 A + 3D ⇄ E
Which of the following correctly applies to this reaction:

Products are favored at equilibrium.

Rate_forward = Rate_reverse

Keq =	[E]
	[A]2[D]3

Keq =	[A]2[D]3
	[E]

C only

B and D only

A, B, and C only

A, B, and D only

For the reaction given below, ΔH = 4.68 kJ.
X(g) + 2 Y(g) ⇄ Z (g)
Which of the following changes will cause a decrease in the concentration of Z?

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For the reaction given below, ΔH = 4.68 kJ.
X(g) + 2 Y(g) ⇄ Z (g)
Which of the following changes will cause a decrease in the concentration of Z?

Increasing the temperature

Increasing the concentration of X

Increasing the size of the container

Addition of a catalyst

Need help with these two questions!

Answer 1

1.

2 A + 3D ⇄ E ; K = 0.00551

So, Keq = [E] / { [A]² [D]³

Since, K < 1. So, the reaction favours the reactant side

Answer is option (C)  Keq = [E] / { [A]² [D]³

2.

X(g) + 2 Y(g) ⇄ Z (g) ; ΔH = 4.68 kJ

Since, ΔH = 4.68 kJ. So, ΔH has a positive value.

Hence, endothermic in forward direction and exothermic in reverse direction.

Decrease in the concentration of Z, means Z reacts to produce X and Y. That is the reverse reaction, which is exothermic reaction.

Exothermic reaction favours when the temperature is decreased.

Again, reactant side = 1 + 2 = 3 moles
product side = 1 mole  

When there is an increase in volume, the equilibrium will shift to favor the direction that produces more moles of gas.

So, answer is Increasing the size of the container.