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Chemical Thermodynamics question

2024 · Q47
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Chemical Thermodynamics question

2024 · Q47

IIT JAMChemistryChemical ThermodynamicsNumerical+1 / −0
The enthalpy change for the reaction C(g)+12O2(g)→CO(g)\mathrm{C}(g) + \frac{1}{2}\mathrm{O}_2(g) \rightarrow \mathrm{CO}(g)C(g)+21​O2​(g)→CO(g) is ‾\underline{\hspace{2cm}}​ kJ per mole of CO(g)\mathrm{CO}(g)CO(g) produced. (rounded off to one decimal place) [Given: C(g)+O2(g)→CO2(g),ΔHrxn=−393.5 kJ per mole of CO2(g) producedCO2(g)→CO(g)+12O2(g),ΔHrxn=283.0 kJ per mole of CO(g) produced]\mathrm{C}(g) + \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g), \Delta\mathrm{H}_{\mathrm{rxn}} = -393.5 \text{ kJ per mole of } \mathrm{CO}_2(g) \text{ produced}\mathrm{CO}_2(g) \rightarrow \mathrm{CO}(g) + \frac{1}{2}\mathrm{O}_2(g), \Delta\mathrm{H}_{\mathrm{rxn}} = 283.0 \text{ kJ per mole of } \mathrm{CO}(g) \text{ produced}]C(g)+O2​(g)→CO2​(g),ΔHrxn​=−393.5 kJ per mole of CO2​(g) producedCO2​(g)→CO(g)+21​O2​(g),ΔHrxn​=283.0 kJ per mole of CO(g) produced]
Numerical answer
View written solutionFree

Correct answer: -111.0, -110.0

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