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.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol