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Step-by-Step Solution
Step 1: Write the balanced combustion reaction
The given reaction for the combustion of ethanol is:
C_{2}H_{5}OH(l) + 3\,O_{2}(g) \to 2\,CO_{2}(g) + 3\,H_{2}O(l).
Step 2: Note the enthalpy change (ΔH)
It is provided that the enthalpy change (heat of combustion) at constant pressure is
ΔH = -327 \text{ kcal}.
Negative sign indicates that the reaction is exothermic (releases heat).
Step 3: Determine the change in moles of gas (Δng)
In the reaction, the number of moles of gaseous reactants is 3 (from 3 O2) and the number of moles of gaseous products is 2 (from 2 CO2). Water is in liquid form, so it does not count toward gaseous mole changes. Thus:
Δn_{g} = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 2 - 3 = -1.
Step 4: Use the relation between ΔH and ΔU
The standard relation between enthalpy change ΔH and internal energy change ΔU is:
ΔH = ΔU + Δn_{g} \, R \, T,
which rearranges to:
ΔU = ΔH - Δn_{g} \, R \, T.
Step 5: Substitute the values and solve for ΔU
Convert ΔH to calories:
-327 \text{ kcal} = -327 \times 1000 \text{ cal} = -327000 \text{ cal}.
Given:
Δn_{g} = -1,
R = 2 \text{ cal mol}^{-1} \text{K}^{-1},
T = 27^\circ \text{C} = 300 \text{ K}.
Therefore,
Δn_{g} \, R \, T = (-1) \times 2 \times 300 = -600 \text{ cal}.
Substituting in the formula:
ΔU = (-327000) - (-600) = -327000 + 600 = -326400 \text{ cal}.
Step 6: Interpret the result
The negative sign indicates an exothermic process. The heat evolved (magnitude) under constant volume is
326400 \text{ cal}.
This is the amount of energy released when the combustion occurs at constant volume.
Final Answer
The heat evolved at constant volume is 326400 cal.
