1. Tardigrade
  2. Question
  3. Chemistry
  4. Calculate the enthalpy change for the reaction, C 2 H 4(g)+ H 2(g) arrow C 2 H 6(g) using the data given below C 2 H 4(g)+3 O 2(g) arrow 2 CO 2(g)+2 H 2 O (l) Δ H =-1415 kJ C 2 H 6(g)+(7/2) O 2(g) arrow 2 CO 2(g)+3 H 2 O Δ H =-1566 kJ H 2(g)+(1/2) O 2(g) arrow H 2 O (l) ; Δ H =-286 kJ

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Q. Calculate the enthalpy change for the reaction,
$C _{2} H _{4}(g)+ H _{2}(g) \rightarrow C _{2} H _{6}(g)$ using the data given below
$C _{2} H _{4}(g)+3 O _{2}(g) \rightarrow 2 CO _{2}(g)+2 H _{2} O (l)$
$\Delta H =-1415\, kJ$
$C _{2} H _{6}(g)+\frac{7}{2} O _{2}(g) \rightarrow 2 CO _{2}(g)+3 H _{2} O$
$\Delta H =-1566\, kJ$
$H _{2}(g)+\frac{1}{2} O _{2}(g) \rightarrow H _{2} O (l) ; \Delta H =-286\, kJ$

J & K CETJ & K CET 2013Thermodynamics

Solution:

According to available data
(i) $C _{2} H _{4}(g)+3 O _{2}(g) \rightarrow 2 CO _{2}(g)+2 H _{2} O (l)$
$\Delta H =-1415\, kJ$
(ii) $C _{2} H _{6}(g)+\frac{7}{2} O _{2}(g) \rightarrow 2 CO _{2}(g)+3 H _{2} O (l)$
$\Delta H =-1566\, kJ$
(iii) $H _{2}(g)+\frac{1}{2} O _{2}(g) \rightarrow H _{2} O (l) ; \Delta H =-286\, kJ$
We aim at $C _{2} H _{4}(g)+ H _{2}(g) \rightarrow C _{2} H _{6}(g) ; \Delta H$
Eq. (i) + Eq. (iii) - Eq. (ii) and the correct
$\Delta H$ value is $=(-1415)+(-286)-(-1566)$
$=-135\, kJ$