Chemical Equilibrium Test

Result

Chemical Equilibrium Test - 5

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Question 1
1 / -0

K_c for a gaseous state reversible reaction is 300 mol^-2 L² at 127^oC. Hence, K_p of the reaction will be

A
300 × (0.082 × 300) atm

B
300 × (8.314 × 400)² atm²

C
300 × (0.082 × 400)^-2 atm^-2

D
300 × (8.314 × 127)² atm²

Solution

K_p = K_c(RT)^Δn ;

Δn = -2 as shown by the units of K_c;

R = 0.082 atm K^-1 mol^-1

Hence,K_p = 300 × (0.082 × 400)^-2 atm^-2
Question 2
1 / -0

Statement - 1 : State of equilibrium of a system cannot be changed by some external factors such as pressure, volume, concentration.

Statement - 2 : Any change in the state of equilibrium caused by externalfactors is nullified by the system.

A
Both Statement – 1 and Statement – 2 are true and Statement – 2 is the correct explanation of Statement – 1.

B
Both Statement – 1 and Statement – 2 are true and Statement – 2 is not the correct explanation of Statement – 1.

C
Statement – 1 is true but Statement – 2 is false.

D
Statement – 1 is false but Statement – 2 is true.

Solution

Assertion is false, reason is true.
There will be a change in state of equilibrium by external factors but there will be no change in the magnitude of equilibrium.
Question 3
1 / -0

The standard free energy changes for the reactions:
2H_2(g) + O_2(g) ⇌ 2H₂O_(g) and CO_(g) + H₂O_(g) ⇌ CO_2(g) + H_2(g) are - 457.0 kJ and - 28.5 kJ respectively. Standard free energy for the reaction: 2CO₂ ⇌ 2 CO_(g) + O_2(g) will be:

A
484.4 kJ

B
– 485.4 kJ

C
– 514.0 kJ

D
514.0 kJ

Solution

On multiplying second eq. by 2 and adding to the first eqn.:

2CO_(g) + O_2(g) ⇌ 2CO_2(g)

ΔG = - 457.0 + 2 × (- 28.5) = - 514.0 kJ

Hence, for 2CO_2(g) ⇌ 2CO_(g) + O_2(g), ΔG = 514.0 kJ