Question

Given the following thermochemical equations, $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow & 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & & \Delta H=-1299.5 \mathrm{~kJ} \\ \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & & \Delta H=-393.5 \mathrm{~kJ} \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & & \Delta H=-285.8 \mathrm{~kJ} \end{aligned}$$ calculate \(\Delta H\) for the decomposition of one mole of acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}(g),\) to its elements in their stable state at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm} .\)

Short answer

The enthalpy change for the decomposition of one mole of acetylene into its constituent elements in their stable state at 25°C and 1 atm is -2491.3 kJ.

Step-by-step solution

Determine the target equation

First, we need to write the target equation, which represents the process of decomposing one mole of acetylene into its constituent elements in their stable state at \(25^{\circ}\mathrm{C}\) and \(1\mathrm{~atm}\). The target equation can be written as: $$ \mathrm{C}_{2}\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{C}(s) + \mathrm{H}_{2}(g) $$

Reverse Equation 3

Since we want to produce hydrogen gas on the right side of our target equation, we have to reverse Equation 3. When reversing a reaction, the sign of \(\Delta H\) must be changed as well. The reversed reaction and the new \(\Delta H\) value become: $$ \mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{H}_{2}(g) + \frac{1}{2}\mathrm{O}_{2}(g) \hspace{15pt} \Delta H = 285.8\mathrm{~kJ} $$

Manipulate Equation 1

In Equation 1, only half a mole of \(\mathrm{C}_{2}\mathrm{H}_{2}(g)\) is consumed. To match the amount of \(\mathrm{C}_{2}\mathrm{H}_{2}\) in the target equation (1 mole), we need to multiply Equation 1 by 2. The modified Equation 1 becomes: $$ 2\mathrm{C}_{2}\mathrm{H}_{2}(g) + 5\mathrm{O}_{2}(g) \longrightarrow 4\mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(l) \hspace{15pt} \Delta H = -2599.0\mathrm{~kJ} $$

Add Equations 2 and 3

We can now add Equation 2 and the reversed Equation 3 together to eliminate one mole of \(\mathrm{CO}_{2}(g)\) and produce one mole of \(\mathrm{H}_{2}(g)\). The sum of these reactions becomes: $$ \mathrm{C}(s) + \mathrm{O}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{CO}_{2}(g) + \mathrm{H}_{2}(g) + \frac{1}{2}\mathrm{O}_{2}(g) \hspace{15pt} \Delta H = -393.5\mathrm{~kJ} + 285.8\mathrm{~kJ} = -107.7\mathrm{~kJ} $$

Subtract the sum from the modified Equation 1

Finally, we will subtract the sum from Step 4 from the modified Equation 1 to eliminate \(\mathrm{CO}_{2}\), \(\mathrm{H}_{2}\mathrm{O}\). and \(\mathrm{O}_{2}\) from the overall reaction. The new overall reaction and enthalpy change will be: $$ \mathrm{C}_{2}\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{C}(s) +\mathrm{H}_{2}(g) \hspace{15pt} \Delta H = -2599.0\mathrm{~kJ} - (-107.7\mathrm{~kJ}) = -2491.3\mathrm{~kJ} $$

Conclusion

The enthalpy change for the decomposition of one mole of acetylene, \(\mathrm{C}_{2}\mathrm{H}_{2}(g)\), into its constituent elements in their stable state at \(25^{\circ}\mathrm{C}\) and \(1\mathrm{~atm}\) is \(\Delta H = -2491.3\mathrm{~kJ}\).