Question

Calculate the standard enthalpy of the reaction,  2C(graphite) + 3H₂(g) →C₂H₆(g) ; ΔH° = ? 

from the following ΔH° values: 

a. C₂H_6(g) + \(\frac72\)O_2(g) → 2CO_2(g) + 3H₂O_(l) ; ΔH° = –1560 kJ 

b. H_2(g) + \(\frac12\) O_2(g) → H₂O_(l) ; ΔH° = –285.8 kJ 

c. C_(graphite) + O_2(g) → CO_2(g) ; ΔH° = –393.5 kJ.

Answer 1

Given: Given equations are,

a. C₂H_6(g) + \(\frac72\)O_2(g) → 2CO_2(g) + 3H₂O_(l), ΔH° = –1560 kJ      .......(1)

b. H_2(g) + \(\frac12\) O_2(g) → H₂O_(l), ΔH° = –285.8 kJ                           ........(2)

c. C_(graphite) + O_2(g) → CO_2(g), ΔH° = –393.5 kJ.                      .........(3)

To find: Standard enthalpy of the given reaction 

Calculation: Reversing equation (1),

2CO_2(g) + 3H₂O_(l) →  C₂H_6(g) + \(\frac72\)O_2(g), ΔH° = –1560 kJ      .......(4)

Multiplying equation (2) by 3 and (3) by 2, then adding to equation (4)

ΔH°  = 1560 + (-857.4) + (-787.0) = -84.4 kJ