JEE Main 2014 · 6 Apr · Shift 1 · Chemistry Q36

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ChemistrySingle correct+4 / -1Question 36 of 90

For complete combustion of ethanol,
$\ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l)},$
the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ mol $^{-1}$ at 25°C. Assuming ideality the Enthalpy of combustion, $\Delta_cH$ , for the reaction will be :

( $R=8.314$ kJ mol $^{-1}$ )

-1366.95

kJ mol

^{-1}

-1361.95

kJ mol

^{-1}

-1460.50

kJ mol

^{-1}

-1350.50

kJ mol

^{-1}

Answer and worked solution

Official answer: Bonus: full marks to everyone

Official key: marks to all.

The bomb calorimeter gives $\Delta U=-1364.47$ kJ mol⁻¹. With $\Delta n_g=2-3=-1$ ,
$\Delta H=\Delta U+\Delta n_gRT=-1364.47-(8.314\times10^{-3})(298)=-1366.95\text{ kJ mol}^{-1},$
option (1). But the paper gives $R=8.314$ kJ mol⁻¹ (instead of J K⁻¹ mol⁻¹). Taken literally, that does not give any option, so marks were awarded to all.

Topics:First law and enthalpy Calorimetry

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JEE Main 2014 · 6 Apr · Shift 1, Chemistry question 36