Question

Calculate the enthalpy change for the reaction,
$C_2{H}_4(g)+H_2(g)\to C_2{H}_6(g)$ using the data given below
$C_2{H}_4(g)+3O_2(g)\to 2C{O}_2(g)+2H_{2}O(l)$
$\Delta H=-1415kJ$
$C_2{H}_6(g)+\frac{7}{2}{O}_2(g)\to 2C{O}_2(g)+3H_{2}O$
$\Delta H=-1566kJ$
$H_2(g)+\frac{1}{2}{O}_2(g)\to H_{2}O(l);\Delta H=-286kJ$

• A. -437 kJ
• B. 135 kJ
• C. -135 kJ
• D. None of these

Answer 1

Answer: (C)

According to available data
(i) $C_2{H}_4(g)+3O_2(g)\to 2C{O}_2(g)+2H_{2}O(l)$
$\Delta H=-1415kJ$
(ii) $C_2{H}_6(g)+\frac{7}{2}{O}_2(g)\to 2C{O}_2(g)+3H_{2}O(l)$
$\Delta H=-1566kJ$
(iii) $H_2(g)+\frac{1}{2}{O}_2(g)\to H_{2}O(l);\Delta H=-286kJ$
We aim at $C_2{H}_4(g)+H_2(g)\to C_2{H}_6(g);\Delta H$
Eq. (i) + Eq. (iii) - Eq. (ii) and the correct
$\Delta H$ value is $=(-1415)+(-286)-(-1566)$
$=-135kJ$