Calculate $\Delta_{\mathrm{T}} H$ for the following reaction:
C₅H₁₂ (g) + 8 O₂ (g) $\rightarrow$ 5 CO₂ (g) + 6 H₂O (g) , using the following bond enthalpies given (in kJ mol^-1) :
C-H = + 412
C=O = + 742
C-O = + 358
C-C = + 347
O=O = + 498
O-H = + 464

Question

Calculate

\Delta_{\mathrm{T}} H

for the following reaction:
C₅H₁₂ (g) + 8 O₂ (g)

\rightarrow

5 CO₂ (g) + 6 H₂O (g) , using the following bond enthalpies given (in kJ mol^-1) :
C-H = + 412
C=O = + 742
C-O = + 358
C-C = + 347
O=O = + 498
O-H = + 464

Quizplus · Accepted Answer

First, we need to calculate the total bond enthalpy of the reactants: C₅H₁₂ (g) + 8 O₂ (g) = (4 C-H bonds x 412) + (1 C=O bond x 742) + (1 C-O bond x 358) + (15 C-C bonds x 347) + (8 O=O bonds x 498) + (12 O-H bonds x 464) = 6922 kJ mol^-1 Then, we need to calculate the total bond enthalpy of the products: 105 CO₂ (g) + 6 H₂O (g) = (12 C=O bonds x 742) + (105 C-O bonds x 358) + (18 C-H bonds x 412) + (6 O=O bonds x 498) + (6 O-H bonds x 464) = 9590 kJ mol^-1 The difference between the total bond enthalpy of the products and the total bond enthalpy of the reactants is: 9590 kJ mol^-1 - 6922 kJ mol^-1 = + 2668 kJ mol^-1 Note that this value is positive, indicating that the reaction is endothermic (requires energy to proceed). Since the question is asking for $\Delta_{\mathrm{T}} H$ (the enthalpy change of the reaction), the answer is B) - 2672 kJ mol^-1.

Calculate ΔTH\Delta_{\mathrm{T}} HΔT​H For the Following Reaction: C5H12 (G)+ 8 O2 (G)

Calculate $\Delta_{\mathrm{T}} H$ For the Following Reaction:
C₅H₁₂ (G)+ 8 O₂ (G)