Question

Evaluate U at 300 K for the following processes:
(i) C2H4 (g) + 3O2 (g) = 2CO2 (g) + 2H2O (l) Given H = 337.3 kCal.
(ii) 2C2H6 (g) + 7O2 (g) = 4CO2 (g) + 6H2O (l) Given H = 745.6 kCal.

Answer 1

Given:

(i) ΔH = - 337.3 k Cal

(ii) ΔH = - 745.6 k Cal

T = 300 K

To Find:

The internal energy of the two systems.

Calculation:

- We know that:

ΔH = ΔU + ΔnRT

⇒ ΔU = ΔH - ΔnRT

- R = 1.98 cal/mol-K

(i) Δn = 2 - 4 = -2

Putting all the values in the equation, we get:

ΔU = - 337.3 - (-2) × 1.98 × 300

⇒ ΔU = - 337.3 + 1188

⇒ ΔU = 850.7 k Cal

(ii) Δn = 4 - 9 = -5

Putting all the values in the equation, we get:

ΔU = - 745.6 - (-5) × 1.98 × 300

⇒ ΔU = - 745.6 + 2970

⇒ ΔU = 2224.4 k Cal

- So, ΔU in first case is 850.7 k Cal and in second case ΔU is 2224.4 k Cal.