NCERT Class XI Chemistry Thermodynamics Solutions
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Question : 14
Total: 22
Calculate the standard enthalpy of formation of C H 3 O H ( l )
C H 3 O H ( l ) +
O 2 ( g ) C O 2 ( g ) + 2 H 2 O ( l ) r H ° k J m o l − 1
C ( graphite ) + O 2 ( g ) C O 2 ( g ) Δ c H ° k J m o l − 1
H 2 ( g ) +
O 2 ( g ) H 2 O ( l ) Δ f H ° k J m o l − 1
Solution:
The given thermochemical equations are
C H 3 O H ( l ) +
O 2 ( g ) C O 2 ( g ) + 2 H 2 O ( l ) Δ r H ° k J m o l − 1
C ( graphite ) + O 2 ( g ) C O 2 ( g ) Δ c H ° k J m o l − 1
H 2 ( g ) +
O 2 ( g ) H 2 O ( l ) Δ f H ° k J m o l − 1
The required thermochemical equation is
C ( s ) + 2 H 2 ( g ) +
O 2 ( g ) C H 3 O H ( l )
2 × eqn. (3) + eqn. (2) – eqn. (1) gives
C ( s ) + 2 H 2 ( g ) +
O 2 ( g ) C H 3 O H ( l ) C ( s ) + 2 H 2 ( g ) +
O 2 ( g ) C H 3 O H ( l )
ΔH = (–286 × 2) + (–393) – (–726) = –572 – 393 + 726 = –239 kJ
∴Δ f H ° C H 3 O H ( l ) k J m o l – 1
The required thermochemical equation is
2 × eqn. (3) + eqn. (2) – eqn. (1) gives
ΔH = (–286 × 2) + (–393) – (–726) = –572 – 393 + 726 = –239 kJ
∴
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