For the following reaction at 300 K A2(g) + 3B2(g) 2AB3(g) the enthalpy change is +15 kJ, then the internal energy change is :
- 19988.4 J
- 200 J
- 1999 J
- 1.9988 kJ
Answer
(1) 19988.4 J
Solution
ng = 2 - (1 + 3) = -2
H = U + ng RT
U = H - ng RT
= 15000 - (-2)(8.314)(300)
= 15000 + 4988.4
= 19988.4 J
Answer: (1)