Chemistry - Thermochemistry Question with Solution | TestHub
ChemistryThermochemistryHess's Law of Constant Heat SummationMedium2 minQB
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Calculate ΔH for the following reaction at 298 K:
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
from the following thermochemical equations:
(i) 3Fe2O3(s) + CO(g) → 2Fe3O4(s) + CO2(g); ΔH = -46.4 kJ
(ii) FeO(s) + CO(g) → Fe(s) + CO2(g); ΔH = 9.0 kJ
(iii) Fe3O4(s) + CO(g) → 3FeO(s) + CO2(g); ΔH = -41.0 kJ
Options:
Answer:
A
Solution:
To achieve the target reaction, 3Fe2O3 + 9CO tarrow 6Fe + 9CO2, we combine the given reactions as follows: (i) + 2(iii) + 6(ii). This combination causes the Fe3O4 and FeO intermediates to cancel out.
The total enthalpy change, 3Δ H, is calculated as:
3Δ H = -46.4 + 2(-41.0) + 6(9.0)
3Δ H = -46.4 - 82.0 + 54.0
3Δ H = -74.4 kJ
Therefore, the enthalpy change for the target reaction, Δ H, is:
Δ H = (-74.4)/3 = -24.8 kJ
This corresponds to option (a).
Stream:JEESubject:ChemistryTopic:ThermochemistrySubtopic:Hess's Law of Constant Heat Summation
⏱ 2mℹ️ Source: QB
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