Animated Solution for Chemistry - Chemical Thermodynamics: If at 298 K, the bond energies of C—H, C—C, C=C and H—H bonds are respectively 414, 347, 615 and 435 kJ mol${}^{-1}$, the value of enthalpy change for the reaction, ${\text{H}}_2\text{C}={\text{CH}}_2(g)+{\text{H}}_2(g)⟶{\text{H}}_3\text{C}—{\text{CH}}_3(g)$ at 298 K will be

Imagine a microscopic battlefield where molecules are torn apart and reassembled into entirely new structures. In this problem, we are witnessing the hydrogenation of ethene. We start with ethene (${\text{H}}_2\text{C}={\text{CH}}_2$) and hydrogen gas (${\text{H}}_2$). Our goal is to transform them into ethane (${\text{H}}_3\text{C}—{\text{CH}}_3$).

To make this transformation happen, we must first invest energy to break the existing bonds in the reactants. This is an endothermic process. Once the atoms are free, they recombine to form new bonds in the products, releasing energy in the process. This is an exothermic event. The overall enthalpy change of the reaction, $\Delta H$, is simply the net balance of this energy exchange.

The fundamental principle of thermochemistry using bond energies is beautifully simple:

Let's take a meticulous inventory of the bonds involved. On the reactant side, ethene possesses one $\text{C}=\text{C}$ double bond and four $\text{C}—\text{H}$ single bonds. The hydrogen molecule adds one $\text{H}—\text{H}$ bond to the mix. On the product side, the newly formed ethane molecule boasts one $\text{C}—\text{C}$ single bond and six $\text{C}—\text{H}$ single bonds.

We could calculate the total energy of all bonds on both sides, but why work harder when we can work smarter? Notice that there are four $\text{C}—\text{H}$ bonds in the reactants and six $\text{C}—\text{H}$ bonds in the products.

Mathematically, we can cancel out the four common $\text{C}—\text{H}$ bonds from both sides. This means the net chemical change is simply breaking one $\text{C}=\text{C}$ double bond and one $\text{H}—\text{H}$ bond, and forming one $\text{C}—\text{C}$ single bond and two new $\text{C}—\text{H}$ bonds. Our simplified equation becomes:

Now, we substitute the given bond energy values into our streamlined equation. The energy required to break the reactant bonds is:

The energy released upon forming the new product bonds is:

Finally, we find the net enthalpy change:

The negative sign is the grand finale. It tells us that the reaction is exothermic. The system released $125\text{ kJ}$ of energy into the surroundings because the newly formed bonds in ethane are collectively stronger and more stable than the original bonds in ethene and hydrogen.