ΔH_Z is the enthalpy change from A+B→E. An alternative route is to go from A+B→C+D and then on from C+D→E. This alternative route 'goes with' the arrows of the enthalpy changes shown so the signs (indicating the exothermic or endothermic nature) of the enthalpy changes will remain unchanged.

Thus the correct answer is ΔH_Z = ΔH_Y + ΔH_X

ΔH_A is the enthalpy change from Z+Y→U. An alternative route is to go from Z+Y→X (against the arrow) and then on from X→V+W (with the arrow) and then on from V+W→U (against the arrow). This alternative route goes against the arrows of the enthalpy changes ΔH_B and ΔH_D so the signs (indicating the exothermic or endothermic nature) of the enthalpy changes will need to be inverted by applying a negaitve sign. However, this alternative route goes with the arrow of the enthalpy changes ΔH_C so the sign (indicating the exothermic or endothermic nature) of this enthalpy change will remain unchanged.

Thus the correct answer is ΔH_Z = − ΔH_B + ΔH_C − ΔH_D

We are given ΔH combustion data so we construct the energy cycle by linking the reactants and products through the combustion products:

The enthalpy change for the reaction given (Y) can then be found: The alternative route going via the combustion products is therefore Y = X−Z (−Z because we are going against the arrow so the sign must be inverted).

Change X is the combustion of two moles of C_(s) and three moles of H_2(g) and change Z is the combustion of one mole of C₂H_6(g) so the calculation is as follows:

X = (2×−394)+(3×−286) = −1646

Z = −1561

Y = X−Z = −1646−(−1561) = −85 kJ mol⁻¹

The correct answer is therefore −85

Incorrect answers

+85 is obtained if the signs of changes X and Z are inverted.

+881 is obtained if the the mole quantities are ignored and one mole of each of C and H₂ are used to obtain change X.

−3207 is obtained if both changes are added together: Y = X + Z

We are given ΔH combustion data so we construct the energy cycle by linking the reactants and products through the combustion products:

The enthalpy change for the reaction given (K) can then be found: The alternative route going via the combustion products is therefore K = J−L (−L because we are going against the arrow so the sign must be inverted).

Change J is the combustion of two moles of C_(s) (we can ignore the O₂(g) since oxygen doesn't combust itself) and change L is the combustion of two moles of CO_(g) so the calculation is as follows:

J = 2×−394 = −788

L = 2×−283 = −566

K = J−L = −788−(−566) = −222 kJ mol⁻¹

The correct answer is therefore −222

Incorrect answers

+222 is obtained if the signs of changes J and L are inverted.

−111 is obtained if the the mole quantities are ignored and one mole of each of C and CO are used.

+111 is obtained if the the mole quantities are ignored and one mole of each of C and CO are used and the the signs of changes J and L are inverted.

We are given ΔH formation data so we construct the energy cycle by linking the reactants and products through the elements:

The enthalpy change for the reaction given (A) can then be found: The alternative route going via the elements is therefore A = −B+C (−B because we are going against the arrow so the sign must be inverted).

Change B is the formation of one mole of C₅H_12(l) (the ΔH formation of any element (O_2(g) here) is by definition zero) and change C is the formation of five moles of CO_2(g) and six moles of H₂O_(l) so the calculation is as follows:

B = −173

C = (5×−393.5)+(6×−285.8) = −3682.3

A = −B+C = −(−173)+(−3682.3) = −3509.3 kJ mol⁻¹

The correct answer is therefore −3509 (4 sig figs)

Incorrect answers

+3509 is obtained if the signs of changes B and C are inverted.

−506.3 is obtained if the the mole quantities are ignored and one mole of each of CO₂ and H₂O are used.

−3855 is obtained if both changes are added together: A = B + C

We are given ΔH formation data so we construct the energy cycle by linking the reactants and products through the elements:

The enthalpy change for the reaction given (E) can then be found: The alternative route going via the elements is therefore E = −F+G (−F because we are going against the arrow so the sign must be inverted).

Change F is the formation of one mole of C₃H_6(g) (the ΔH formation of any element (O_2(g) here) is by definition zero) and change G is the formation of three moles of CO_2(g) and three moles of H₂O_(l) so the calculation is as follows:

F = +20.0

G = (3×−393.5)+(3×−285.8) = −2037.9

E = −F+G = −(+20.0)+(−2037.9) = −2057.9 kJ mol⁻¹

The correct answer is therefore −2058 (4 sig figs)

Incorrect answers

+2058 is obtained if the signs of changes F and G are inverted.

−699.3 is obtained if the the mole quantities are ignored and one mole of each of CO₂ and H₂O are used.

−2018 is obtained if both changes are added together: E = F + G

We are given ΔH data that shows reactions with oxygen (probably combustion) so we construct the energy cycle by linking the reactants and products through the combustion products:

The enthalpy change for the reaction given ('A') can then be found: The alternative route going via the combustion products is therefore A = B−C (−C because we are going against the arrow so the sign must be inverted).

Change 'B' is the combustion of one mole of C_(s) (reaction iii) and one quarter of a mole of S_8(S) (so one quarter of the value of reaction i) and change 'C' is the combustion of one mole of CS_2(g) as given in reaction ii. So the calculation is as follows:

B = −394+(¼×−2376) = −988

C = −1072

A = B−C = −988−(−1072) = +84 kJ mol⁻¹

The correct answer is therefore +84

Incorrect answers

−1304 is obtained if A = i − ii is taken to be the answer.

−1698 is obtained if one quarter of the value for reaction i is not taken (the full value of −2376 is used and all other calculations are correct).

−2060 is obtained if the values for B and C are added together (A = B + C)