Butane (C4 H10(g), Delta. Hf = â€"125. 6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Delta. Hf = â€"393. 5 kJ/mol ) and water (H2 O, Delta. Hf = â€"241. 82 kJ/mol) according to the equation below. 2 upper C subscript 4 upper H subscript 10 (g) plus 13 upper o subscript 2 (g) right arrow 8 upper C upper O subscript 2 (g 0 plus 10 upper H subscript 2 upper O (g). What is the enthalpy of combustion (per mole) of C4H10 (g)? Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants. â€"2,657. 5 kJ/mol â€"5315. 0 kJ/mol â€"509. 7 kJ/mol â€"254. 8 kJ/mol.

Question

contestada

Respuesta :

pstnonsonjoku pstnonsonjoku · Answer 1 · 2021-12-15T11:20:24+01:00

The enthalpy of this reaction is -5315 KJ/mol.

The equation of the reaction is;

2C4H10(g) + 13O2(g) -----> 8CO2 (g) + 10H2O(g)

We know that the enthalpy of reaction can be obtained from the enthalpy of formation of the reactants and products as follows;

ΔHrxn = ΔHf(products) - ΔHf(reactants)

We have the following information from the question;

ΔHf C4H10 = - 125. 6 kJ/mol

ΔHf CO2 = - 393. 5 kJ/mol

ΔHf H2O = - 241. 82 kJ/mol

ΔHf O2 = 0 KJ/mol

Hence;

[(8 × (- 393. 5 )) + (10 × (- 241. 82))] - [2( - 125. 6))]

= -5315 KJ/mol

Learn more: https://brainly.com/question/13164491