JEE MAIN - Chemistry (2003 - No. 6)
If at 298 K the bond energies of C - H, C - C, C = C and H - H bonds are respectively 414, 347, 615 and 435 kJ/mol, the value of enthalpy change for the reaction
H2C = CH2(g) + H2(g) $$\to$$ H3C - CH3(g) at 298 K will be :
H2C = CH2(g) + H2(g) $$\to$$ H3C - CH3(g) at 298 K will be :
- 250 kJ
+ 125 kJ
- 125 kJ
+ 250 kJ
Explanation
$$C{H_2} = C{H_2}\left( g \right) + {H_2}\left( g \right) \to C{H_3} - C{H_3}$$
Enthalpy change $$=$$ Bond energy of reactants $$-$$ Bond energy of products.
$$\Delta H = 1\left( {C = C} \right) + 4\left( {C - H} \right) + 1\left( {H - H} \right) - $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\left( {C - C} \right) - 6\left( {C - H} \right)$$
$$ = 1\left( {C = C} \right) + 1\left( {H - H} \right) - 1\left( {C - C} \right) - 2\left( {C - H} \right)$$
$$ = 615 + 435 - 347 - 2 \times 414$$
$$ = 1050 - 1175$$
$$ = - 125\,kJ.$$
Enthalpy change $$=$$ Bond energy of reactants $$-$$ Bond energy of products.
$$\Delta H = 1\left( {C = C} \right) + 4\left( {C - H} \right) + 1\left( {H - H} \right) - $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\left( {C - C} \right) - 6\left( {C - H} \right)$$
$$ = 1\left( {C = C} \right) + 1\left( {H - H} \right) - 1\left( {C - C} \right) - 2\left( {C - H} \right)$$
$$ = 615 + 435 - 347 - 2 \times 414$$
$$ = 1050 - 1175$$
$$ = - 125\,kJ.$$
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