Question

Given the following data: $$\begin{array}{ll}\mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H=-1300 . \mathrm{kJ} \\\\\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) & \Delta H=-394 \mathrm{~kJ} \\\\\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta H=-286 \mathrm{~kJ}\end{array}$$ Calculate \(\Delta H\) for the reaction $$2 \mathrm{C}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)$$

Short answer

The enthalpy change for the target reaction \(2 \mathrm{C}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)\) is \(\Delta H = 2098 \mathrm{~kJ}\).

Step-by-step solution

Write down the given reactions

Reaction 1: \(\phantom{xx}\mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\phantom{xx}\) \(\Delta H_1=-1300 \mathrm{~kJ}\) Reaction 2: \(\phantom{xx}\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)\phantom{xx}\) \(\Delta H_2=-394 \mathrm{~kJ}\) Reaction 3: \(\phantom{xx}\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)\phantom{xx}\) \(\Delta H_3=-286 \mathrm{~kJ}\) Target Reaction: \(\phantom{xx}2 \mathrm{C}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)\phantom{xx}\) \(\Delta H=?\)

Manipulate given reactions to match the target reaction

Manipulate Reaction 1: Reverse and multiply by 2 \(-2 \times \mathrm{C}_{2} \mathrm{H}_{2}(g)+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\Delta H_1'= 2600 \mathrm{~kJ}\) Manipulate Reaction 2: Multiply by 2 \(2 \times \mathrm{C}(s)+2 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)\phantom{xx}\) \(\Delta H_2'= -788 \mathrm{~kJ}\) Manipulate Reaction 3: Reverse \(\phantom{xx} \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)\phantom{xx}\) \(\Delta H_3'= 286 \mathrm{~kJ}\)

Combine manipulated reactions to obtain the target reaction

Add manipulated Reaction 1, Reaction 2, and Reaction 3: \[-2 \times \mathrm{C}_{2} \mathrm{H}_{2}(g) +5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{CO}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(l)\] \[2 \times \mathrm{C}(s) + 2 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)\phantom{xx}\] \[\phantom{xx} \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{2}(g) + \frac{1}{2} \mathrm{O}_{2}(g)\phantom{xx}\] \[\underline{\phantom{xxxxxxx}+\phantom{xxxxxxxxxxxxxxxxxxxxxxx\nearrow}\phantom{xxxxxxxxxxxxxxxxx+}\phantom{xxxxxxxxxxxxxxxxxxxxxxxxxxxx}\phantom{xx}}\] \[2 \mathrm{C}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)\]

Calculate the ΔH for the target reaction

Sum the enthalpy changes of the manipulated reactions to find the enthalpy change of the target reaction: \[\Delta H_{target} = \Delta H_1' + \Delta H_2' + \Delta H_3' = 2600 \mathrm{~kJ} - 788 \mathrm{~kJ} + 286 \mathrm{~kJ} = 2098 \mathrm{~kJ}\] Therefore, the enthalpy change for the target reaction is ΔH = 2098 kJ.

Key concepts

Enthalpy Change

Enthalpy change is a fundamental concept in chemistry that describes the heat exchange in a system during a chemical reaction. It is often represented by the symbol \( \Delta H \).
When a reaction releases heat, it is called exothermic and \( \Delta H \) is negative, indicating energy is released to the surroundings. Conversely, if a reaction absorbs heat, it is endothermic and \( \Delta H \) is positive, meaning energy is absorbed from the surroundings.
The enthalpy change helps us understand the energy dynamics of reactions. By using Hess's Law, which states that the total enthalpy change is the same no matter the pathway taken, we can calculate unknown enthalpy changes. This is done by combining other known reactions, making enthalpy an essential part of energy calculations in chemistry.

Chemical Reactions

A chemical reaction involves transforming one or more substances into new substances with different properties. This transformation occurs through breaking and forming bonds, which inherently involve changes in energy.
Each specific reaction is associated with a characteristic enthalpy change that reflects the energy required to break bonds and the energy released from forming new ones. These reactions are balanced according to the principle of conservation of mass.

• Reactants are the starting substances.
• Products are the new substances formed.

In multi-step reactions, Hess's law allows us to add known enthalpies for intermediate reactions to find the total enthalpy change. This step-by-step process helps chemists design efficient reaction pathways by minimizing energy costs.

Energy Calculations

Energy calculations play a crucial role in understanding chemical reactions. They allow us to determine the feasibility and efficiency of reactions by calculating the enthalpy changes.
Using Hess's Law, we can manipulate and combine different reactions to determine the total enthalpy change for a process. This involves reversing reactions or multiplying them by coefficients to match the desired chemical equation.

• Reversing a reaction changes the sign of \( \Delta H \).
• Multiplying a reaction by a number multiplies \( \Delta H \) by the same factor.

By performing these calculations, chemists can decide which conditions are most favorable for a reaction, making energy calculations vital for practical applications in industry and research.