Consider the following system at equilibrium where H° = -111 kJ, and Kc = 0.159 , at 723 K: N2 (g) + 3 H2 (g) 2 NH3 (g)

If the TEMPERATURE on the equilibrium system is suddenly decreased :

The value of Kc A. Increases B. Decreases C. Remains the same

The value of QcA. Is greater than Kc B. Is equal to Kc C. Is less than Kc

The reaction must: A. Run in the forward direction to restablish equilibrium. B. Run in the reverse direction to restablish equilibrium. C. Remain the same. Already at equilibrium.

The concentration of H2 will: A. Increase. B. Decrease. C. Remain the same.

The value of will increase.

The value of will decrease.

The reaction must run in the forward direction to re-establish equilibrium.

The concentration of hydrogen gas will decrease.

Explanation:

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.

As, enthalpy of the reaction is negative. So, it is an exothermic reaction.

For an exothermic reaction, heat is released during a chemical reaction and is written on the product side.

As, heat is released during a chemical reaction. This means that temperature is decreased on the reactant side. If the temperature in the equilibrium is decreased, the equilibrium will shift in the direction where, temperature is getting increased.

Thus, the reaction will shift in right direction that is towards the product to reestablish the equilibrium..

As the reaction will move in forward direction the value of will increase and the value of will decrease.

( Move in froward direction )

Since, the equilibrium will be shifting in forward direction the concentration of hydrogen gas will decrease.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2