Question

1.) jika diketahui deltaH°f CO₂(g)= -394 kJ/mol, deltaH°f CO(g)= -110 kJ/mol, deltaH°f H₂O(G)=-286kJ/mol, deltaH°f NH₃= -46,19kJ/mol, deltaH°f NO= 90,37kJ/mol, hitunglah perubahan entalpi reaksi pada reaksi berikut ini:
a. CO₂ + H₂ --> CO + H₂O
b. 4NH₃ + 5O₂ --> 4NO + 6H₂O

Answer 1

A) CO2 + H2 ----> CO + H2O

O=C=O + H-H ----> C=O + H-O-H

deltaHr = (2EC=O + EH-H) - (EC=O + 2EH-O )
deltaHr =( 2(715)+435 ) - (715 + 2(463))
deltaHr = (1430+435) - (715+926)
deltaHr = 1865 - 1640
deltaHr = 225

Answer 2

A) CO2 + H2 ----> CO + H2O

O=C=O + H-H ----> C=O + H-O-H

deltaHr = (2EC=O + EH-H) - (EC=O + 2EH-O )
deltaHr =( 2(715)+435 ) - (715 + 2(463))
deltaHr = (1430+435) - (715+926)
deltaHr = 1865 - 1640
deltaHr = 225