Question

Study the reactions for the formation of compounds from their elements. I. C(s) + O2(g) → CO2(g) ΔHf = −394 kJ II. H2(g) + 12O2(g) → H2O(l) ΔHf = −242 kJ III. 2C(s) + 3H2(g) → C2H6(g) ΔH =−84 kJ The combustion of C2H6 is shown by the following equation: C2H6(g) + 72O2(g) → 2CO2(g) + 3H2O(l) Which option correctly gives the enthalpy of combustion of 0.2 moles of C2H6(g)? −1,430 kJ 286 kJ −286 kJ 1,430 kJ Exam 3 Click on the numbers to jump from one question to another. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Answer 1

Positive 1,430

Step-by-step explanation:

Answer 2

The correct option is -286 kJ

Step-by-step explanation:

The given parameters are

C(s) + O₂(g) → CO₂(g) ΔHf = -394 kJ

H₂(g) + 12O₂(g)→H₂O ΔHf = -242 kJ

2C(s) + 3H₂(g)→C₂H₆(g) ΔH = -84 kJ

Te given equation is C₂H₆(g) + 7/2O₂(g) →2CO₂(g) + 3H₂O(l)

The heat of formation or enthalpy of combustion = Heat of formation of the products - Heat of formation of the reactants

The enthalpy of combustion of the reaction = 2*(-394) + 3*(-242)- ((-84)+7/2*0)) = -1,430 kJ

Given that the reaction consists of one mole of C₂H₆(g), we have;

The enthalpy of combustion of one mole of C₂H₆(g) = -1,430 kJ

Therefore, the enthalpy of combustion of 0.2 mole of C₂H₆(g) = -1,430 kJ × 0.2 = -286 kJ

The correct option = -286 kJ.