Calculate the standard enthalpy change, in kJ mol–1, for the reaction between carbon disulfide, CS2, and oxygen shown in the following equation.
The Correct Answer and Explanation is:
Sure! First, we need the balanced chemical equation for the combustion of carbon disulfide (CS₂) in oxygen: CS2(l)+3O2(g)→CO2(g)+2SO2(g)\text{CS}_2(l) + 3\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{SO}_2(g)
Step 1: Use Standard Enthalpies of Formation (ΔH⁰f)
To calculate the standard enthalpy change of reaction (ΔH⁰rxn), use the formula: ΔHrxn∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H^\circ_{\text{rxn}} = \sum \Delta H^\circ_f(\text{products}) – \sum \Delta H^\circ_f(\text{reactants})
Standard enthalpies of formation (in kJ/mol):
- ΔH⁰f [CS₂(l)] = +116.9
- ΔH⁰f [O₂(g)] = 0 (element in standard state)
- ΔH⁰f [CO₂(g)] = –393.5
- ΔH⁰f [SO₂(g)] = –296.8
Step 2: Plug into the Formula
ΔHrxn∘=[1(−393.5)+2(−296.8)]−[1(116.9)+3(0)]\Delta H^\circ_{\text{rxn}} = [1(-393.5) + 2(-296.8)] – [1(116.9) + 3(0)] =[−393.5−593.6]−[116.9]= [-393.5 – 593.6] – [116.9] =−987.1−116.9=−1104.0 kJ mol−1= -987.1 – 116.9 = \boxed{-1104.0\ \text{kJ mol}^{-1}}
Explanation
The standard enthalpy change of a chemical reaction, denoted as ΔH⁰rxn, represents the heat energy released or absorbed when a reaction occurs under standard conditions (298 K and 1 atm pressure), with all reactants and products in their standard states. In this case, we examine the combustion of carbon disulfide (CS₂), a flammable liquid, in the presence of oxygen to form carbon dioxide (CO₂) and sulfur dioxide (SO₂).
To determine the enthalpy change, we apply Hess’s Law, which states that the total enthalpy change of a reaction depends only on the enthalpy of the products and the reactants—not the path taken. Therefore, we can use tabulated standard enthalpies of formation (ΔH⁰f), which reflect the enthalpy change when one mole of a compound forms from its elements in their standard states.
The formula used is: ΔHrxn∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H^\circ_{\text{rxn}} = \sum \Delta H^\circ_f(\text{products}) – \sum \Delta H^\circ_f(\text{reactants})
Here, CS₂ reacts with three moles of O₂ to produce one mole of CO₂ and two moles of SO₂. Using standard values, we substitute and calculate that the reaction has an enthalpy change of –1104.0 kJ/mol, indicating it is strongly exothermic. This makes sense because combustion reactions typically release energy as bonds in reactants are broken and more stable bonds form in the products.
Understanding ΔH⁰rxn is crucial in thermodynamics, environmental science, and industrial applications, where energy balances and heat generation are important for safety, efficiency, and sustainability.
