Question

Use the data below to calculate the standard enthalpy change of formation of Bi(CH3)3() in kJ mol¹

$\Delta_c H^\circ[Bi (CH_3)_3(()] = -2912 kJ mol^{-1}$; $\Delta_f H^\circ[CO_2(g)] = -394 kJ mol^{-1}$

$\Delta_f H^\circ[H_2O(()] = -286 kJ mol^{-1}$; $\Delta_f H^\circ[Bi_2O_3(s)] = -574 kJ mol^{-1}$

Answer 1

The standard enthalpy change of formation of Bi(CH₃)₃(l) is +156 kJ/mol.

Answer 2

1. Write the balanced combustion reaction for one mole of Bi(CH₃)₃. First, balance using 2 moles and then divide by 2:

   2 Bi(CH₃)₃(l) + 12 O₂(g) → Bi₂O₃(s) + 6 CO₂(g) + 9 H₂O(l)

   For one mole:

   Bi(CH₃)₃(l) + 6 O₂(g) → ½ Bi₂O₃(s) + 3 CO₂(g) + (9/2) H₂O(l)

2. Use Hess's law with the combustion equation:

   Δ_cH° = [0.5·Δ_fH°(Bi₂O₃) + 3·Δ_fH°(CO₂) + (9/2)·Δ_fH°(H₂O)] – Δ_fH°(Bi(CH₃)₃)

   Given:

   Δ_cH°[Bi(CH₃)₃] = –2912 kJ/mol
   Δ_fH°(Bi₂O₃) = –574 kJ/mol
   Δ_fH°(CO₂) = –394 kJ/mol
   Δ_fH°(H₂O) = –286 kJ/mol

3. Calculate the total enthalpy for products:

   0.5·(–574) = –287 kJ
   3·(–394) = –1182 kJ
   (9/2)·(–286) = –4.5×286 = –1287 kJ
   Sum = –287 – 1182 – 1287 = –2756 kJ

4. Plug into the equation:

   –2912 = (–2756) – Δ_fH°(Bi(CH₃)₃)

   Solve for Δ_fH°(Bi(CH₃)₃):

   Δ_fH°(Bi(CH₃)₃) = –2756 + 2912 = +156 kJ/mol