Question

5a. Hydrogen cyanide is produced in the following balanced equation: 2NH3 (g) + 3O2 (g) + 2 CH4 (g) → 2HCN (g) + 6 H2O (g). Given the standard heats of formation: ∆H o f (kJ/mol) NH3 (g) - 46, CH4 (g) -75, HCN (g) 135, H2O (g) -242 Calculate the approximate H of this reaction.

5b. Given: Cu2O (s) + ½ O2 (g) → 2 CuO (s) H = -144 kJ

2Cu2O (s) → 2Cu(s) + 2CuO (s) H = 22 kJ Calculate the standard enthalpy of formation of CuO (s

Answer 1

a) ∆Hrxn = 2∆Hf(HCN) + 6∆Hf(H2O) - 2∆Hf(NH3) - 3∆Hf(O2) - 2∆Hf(CH4)

∆Hrxn = 2(135)+6(-242)-2(-46)-3(0)-2(-75) KJ

= 270-1452+92+150 KJ = -940 KJ

b) Multiply equation 1 and reversing equation 2 we get :

2Cu2O + O2 = 4CuO ∆H = -288 KJ

2CuO + 2Cu = 2Cu2O ∆H = -22 KJ

Adding both above equation :

2Cu + O2 = 2CuO ∆H = -310 KJ

For one mole of formation of CuO, ∆H = -310/2 = -155 KJ