Question

estimate the enthalpy change for the reaction ( c_{2}h_{4}(g)+h_{2}(g)\to c_{2}h_{6}(g) ) given the following bond energies.

bond	h - h	c - h	c - c	c = c
bond energy (kj/mol)	436	414	347	620

• 392 kj

+119 kj

• 119 kj

+152 kj
+392 kj

Explanation:

Step1: Calculate energy for breaking bonds

Energy for breaking bonds: \( (4\times414 + 620+436)\text{ kJ/mol}\)

$$\begin{align*} &4\times414+620 + 436\\ =&1656+620+436\\ =&2712\text{ kJ/mol} \end{align*}$$

Step2: Calculate energy for forming bonds

Energy for forming bonds: \( (6\times414+347)\text{ kJ/mol}\)

$$\begin{align*} &6\times414 + 347\\ =&2484+347\\ =&2831\text{ kJ/mol} \end{align*}$$

Step3: Calculate enthalpy change

Enthalpy change \(\Delta H=\) (Energy for breaking bonds) - (Energy for forming bonds)
\(\Delta H=2712 - 2831=- 119\text{ kJ/mol}\)

Answer:

-119 kJ