During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for

NCERT Exemplar · Q5

Q. During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is

(i) 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l) ΔcH = -2658.0 kJ mol^-1

(ii) C4H10(g) + (13/2) O2(g) → 4CO2(g) + 5H2O(g) ΔcH = -1329.0 kJ mol^-1

(iii) C4H10(g) + (13/2) O2(g) → 4CO2(g) + 5H2O(l) ΔcH = -2658.0 kJ mol^-1

(iv) C4H10(g) + (13/2) O2(g) → 4CO2(g) + 5H2O(l) ΔcH = +2658.0 kJ mol^-1

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Concept understanding — Enthalpy and Types of Enthalpy Change

Enthalpy, $H$ , is a state function defined as $H = U + PV$ , introduced because the overwhelming majority of real chemical reactions take place in open vessels at essentially constant atmospheric pressure, rather than at constant volume. At constant pressure, the change in enthalpy equals the heat exchanged with the surroundings, $\Delta H = q_p$ — precisely the quantity a calorimeter measures for a reaction carried out in an open flask, container, or solution. Enthalpy and internal energy are related, for gas-phase reactions at constant temperature, through $\Delta H = \Delta U + \Delta n_g RT$ , where $\Delta n_g$ is the change in moles of gas across the reaction.

Chemists have defined a family of precisely standardized enthalpy changes, each describing one specific, carefully defined process, denoted by the standard-state superscript $^\circ$ (usually $298\ \text{K}$ , $1\ \text{bar}$ ):

Standard enthalpy of formation ( $\Delta_fH^\circ$ ) — one mole of a compound formed from its elements in their standard states; by convention, an element in its standard state has $\Delta_fH^\circ = 0$ .
Standard enthalpy of combustion ( $\Delta_cH^\circ$ ) — one mole of a substance completely burnt in excess oxygen; always exothermic (negative), and readily measured by bomb calorimetry.
Standard enthalpy of atomization ( $\Delta_aH^\circ$ ) — one mole of a substance completely converted into gaseous atoms; always endothermic (positive).
Bond dissociation enthalpy — the energy to break one mole of a specific covalent bond in a gaseous molecule; dividing an atomization enthalpy by the number of equivalent bonds gives an average bond enthalpy.
Enthalpy of ionization (first, second, …) — the energy to remove one mole of electrons from one mole of gaseous atoms (first) or from the resulting gaseous cations (second, always larger than the first, since removing an electron from an already-positive species is harder). …

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A thermochemical equation must show the combustion of one mole of butane (since $\Delta_c H$ is per mole of fuel), produce liquid water (standard combustion convention), and carry a negative sign for the exothermic release of 2658 kJ.

Why thermochemical equations encode both stoichiometry and energy

A thermochemical equation is not just a balanced chemical equation—it is a statement that ties a specific amount of substance to a specific energy change. The enthalpy of combustion $\Delta_c H$ is defined as the enthalpy change when one mole of the substance burns completely in oxygen. The sign convention is crucial: combustion releases heat to the surroundings, so $\Delta_c H$ is negative (the system loses energy). Finally, standard combustion enthalpies are measured with products in their most stable states at 298 K and 1 bar—for water, that means the liquid phase.

Let's decode each option against these principles.

Step-by-step elimination

Check the stoichiometry of butane. The question states "one mole of butane" releases 2658 kJ. Option (i) writes

2\text{C}_4\text{H}_{10}(g) + 13\text{O}_2(g) \to 8\text{CO}_2(g) + 10\text{H}_2\text{O}(l)

This equation burns two moles of butane. If the enthalpy change is written as $-2658.0 \text{ kJ mol}^{-1}$ , it implies 2658 kJ per mole of reaction as written—but that reaction consumes two moles of butane, so per mole of butane the heat released would be only 1329 kJ. The label is inconsistent with the stoichiometry.

Option (i) is incorrect.

Examine the physical state of water. Standard combustion data conventionally report water as liquid $\text{H}_2\text{O}(l)$ because that is the stable phase at 298 K. Option (ii) shows

\text{C}_4\text{H}_{10}(g) + \frac{13}{2}\text{O}_2(g) \to 4\text{CO}_2(g) + 5\text{H}_2\text{O}(g)

with $\Delta_c H = -1329.0 \text{ kJ mol}^{-1}$ . Even ignoring the magnitude for a moment, gaseous water is not the standard state. Moreover, 1329 kJ is exactly half of 2658 kJ—this would be correct only if we had half a mole of butane or if the given 2658 kJ applied to two moles.

Option (ii) is incorrect on both the state of water and the enthalpy value.

Check the sign of $\Delta_c H$ . Combustion is exothermic: the system releases energy, so $\Delta H < 0$ . Option (iv) writes …

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Q.During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is

Why thermochemical equations encode both stoichiometry and energy

Step-by-step elimination

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