Question

Use the given heats of formation to calculate the enthalpy change for this reaction. B2O3(g) + 3COCl2(g) →2BCl3(g) + 3CO2(g) ΔHof of B2O3(g) is –1272.8 kJ∙mol–1 ΔHof of BCl3(g) is –403.8 kJ∙mol–1 ΔHof of COCl2(g) is –218.8 kJ∙mol–1 ΔHof of CO2(g) is –393.5 kJ∙mol–1 Question 8 options: 1) 649.3 kJ·mol–1 2) 354.9 kJ·mol–1 3) –58.9 kJ·mol–1 4) –3917.3 kJ·mol–1

Answer 1

Answer:

the enthalpy change for this reaction is -57.7 kJ/mol

Explanation:

Given:

HB₂O₃ = -1272.8 kJ/mol

HCOCl₂ = -218.8 kJ/mol

HBCl₃ = -403.8 kJ/mol

HCO₂ = -393.5 kJ/mol

Those are all standard enthalpies

Question: Calculate the enthalpy change for this reaction, ΔHreaction = ?

The enthalpy of the reaction is calculated using the standard enthalpies of formation of both products and reagents. To understand better, the reaction is as follows

B₂O₃ + 3COCl₂ → 2BCl₃ + 3CO₂

Where the compounds on the left are the reactants and the compounds on the right are the products

ΔHreaction = ∑ΔHproducts - ∑ΔHreactants

$delta(H)_{reaction} =((2*(-403.2)+(3*(-393.5))-((1*(-1272.8)+(3*(-218.8))=-57.7kJ/mol$

Please be careful with the signs.