JEE Main 2023 · Chemistry · Chemical Thermodynamics · medium

For the combustion C2H4(g)+3O2(g)→2CO2(g)+2H2O(l) in a bomb calorimeter (heat capacity 5.00 kJ K^-1) at 300 K, burning 0.280 g of ethene (M=28)…

Numerical answer

For the combustion

\mathrm{C_2H_4(g)+3O_2(g)\to2CO_2(g)+2H_2O(l)}

in a bomb calorimeter (heat capacity

5.00\ \mathrm{kJ\,K^{-1}}

) at

300\ \mathrm K

, burning

0.280\ \mathrm g

of ethene (

M=28

) raises the temperature by

2.82\ \mathrm K

. The magnitude of the enthalpy of combustion of ethene,

|\Delta_cH|

, in

\mathrm{kJ\,mol^{-1}}

to the nearest integer is (

R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}

)

Show answer and solution

Answer: 1415

Heat released

=5.00\times2.82=14.1

kJ for

0.280/28=0.0100

mol, so

\Delta U=-1410\ \mathrm{kJ\,mol^{-1}}

\Delta n_g=2-4=-2

(liquid water has no gas moles).

\Delta H=\Delta U+\Delta n_gRT=-1410+(-2)(8.314)(300)/1000=-1410-4.99=-1415\ \mathrm{kJ\,mol^{-1}}

. So

|\Delta_cH|\approx1415

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