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Chemical Thermodynamics question

2023 · Q44
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Chemical Thermodynamics question

2023 · Q44

IIT JAMChemistryChemical ThermodynamicsNumerical+1 / −0
Consider the following reaction: 2C6H6+15O2→12CO2+6H2OΔrH2980=−3120 kJ mol−1.2 \mathrm{C}_{6} \mathrm{H}_{6} + 15 \mathrm{O}_{2} \rightarrow 12 \mathrm{CO}_{2} + 6 \mathrm{H}_{2} \mathrm{O} \quad \Delta_{\mathrm{r}} H_{298}^{0} = -3120 \mathrm{~kJ} \mathrm{~mol}^{-1}.2C6​H6​+15O2​→12CO2​+6H2​OΔr​H2980​=−3120 kJ mol−1. A closed system initially contains 5 moles of benzene and 25 moles of oxygen under standard conditions at 298 K. The reaction was stopped when 17.5 moles of oxygen is left. The amount of heat evolved during the reaction is ‾\underline{\hspace{2cm}}​ kJ. (round off to the nearest integer)
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Correct answer: -3120

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