Question

Consider the following system at equilibrium where AH° = -10.4 kJ, and K. = 55.6, at 698 K. H2(E) +13(g) = 2HIE) If the VOLUME of the equilibrium system is suddenly decreased at constant temperature: The value of K. A. increases. B. decreases C. remains the same. The value of QC A. is greater than Kc B. is equal to Ke C. is less than K The reaction must: A. run in the forward direction to reestablish equilibrium. B. run in the reverse direction to reestablish equilibrium. C. remain the same. It is already at equilibrium. The number of moles of 12 will: A. increase. B. decrease C. remain the same.

Answer 1

Answer:

For 1: The correct answer is Option C.

For 2: The correct answer is Option B.

For 3: The correct answer is Option C.

For 4: The correct answer is Option C.

Explanation:

We are given:

$K_e=55.6$

$\Delta H^o_{rxn}=-10.4kJ$

For the given chemical reaction:

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Any change in volume during equilibrium is studied on the basis of change of moles of reactants and products.

When volume is decreased, the equilibrium will shift in the direction which produces fewer moles of gas and when volume is increased, the equilibrium will shift in the direction which produces more moles of gas.

Number of moles of gases on reactant side = [1 + 1] = 2

Number of moles of gases on product side = 2

Change in number of moles = 2 -2 = 0

As, number of moles on both the side of the reaction is same. This means there will be no effect on change in volume.

• For 1:

As, there is no effect of volume on equilibrium. So, the equilibrium constant remains the same.

Hence, the correct answer is Option C.

• For 2:

$K_c$ is the constant of a certain reaction at equilibrium while $Q_c$ is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

There are 3 conditions:

• When $K_{e}>Q_c$; the reaction is product favored.
• When $K_{e}$; the reaction is reactant favored.
• When $K_{e}=Q_c$; the reaction is in equilibrium.

As, the value of $K_e$ is not changing. This means that value of $Q_c$ is will also remain the same.

Hence, the correct answer is Option B.

• For 3:

As, the equilibrium constant is not changing. So, the reaction is present at equilibrium.

Hence, the correct answer is Option C.

• For 4:

The reaction is present at equilibrium. So, the concentration of iodine will remain the same.

Hence, the correct answer is Option C.