Question

The enthalpy of combustion values of $C_{2}H_{4(g)}$, $C_{(graphite,s)}$ and $H_{2(g)}$ are respectively $-1411\ \text{kJ mol}^{-1}$, $-394\ \text{kJ mol}^{-1}$ and $-286\ \text{kJ mol}^{-1}$. What is the enthalpy of formation of $C_{2}H_{4(g)}$ in $\text{kJ mol}^{-1}$?

• A. -102
• B. -51
• C. +102
• D. +153
• E. +51

Hint:

Remember to multiply enthalpy values by their stoichiometric coefficients.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
This problem requires the use of Hess's Law to calculate the enthalpy of formation (\( \Delta H_f^\circ \)) of a compound from given enthalpies of combustion (\( \Delta H_c^\circ \)).
Step 2: Key Formula or Approach:
1. Target Reaction: Write the balanced chemical equation for the formation of C\(_2\)H\(_4\) from its elements in their standard states. \[ 2\text{C}_{(graphite)} + 2\text{H}_{2(g)} \rightarrow \text{C}_2\text{H}_{4(g)} \quad \Delta H_f^\circ = ? \] 2. Given Reactions: Write the balanced chemical equations for the combustion of each substance. (i) C\(_2\)H\(_{4(g)} + 3\text{O}_{2(g)} \rightarrow 2\text{CO}_{2(g)} + 2\text{H}_2\text{O}_{(l)}\), \( \Delta H_c^\circ = -1411 \) kJ/mol (ii) C\(_{(graphite)} + \text{O}_{2(g)} \rightarrow \text{CO}_{2(g)}\), \( \Delta H_c^\circ = -394 \) kJ/mol (iii) H\(_{2(g)} + \frac{1}{2}\text{O}_{2(g)} \rightarrow \text{H}_2\text{O}_{(l)}\), \( \Delta H_c^\circ = -286 \) kJ/mol 3. Hess's Law: Manipulate the given reactions (reverse, multiply by coefficients) so that they add up to the target reaction. The enthalpy changes are manipulated in the same way. Alternatively, use the general formula: \[ \Delta H_{reaction}^\circ = \sum (\Delta H_f^\circ \text{products}) - \sum (\Delta H_f^\circ \text{reactants}) \] For a combustion reaction, this becomes: \[ \Delta H_c^\circ = \sum (\Delta H_f^\circ \text{products}) - \sum (\Delta H_f^\circ \text{reactants}) \] Step 3: Detailed Explanation:
Let's use the method of manipulating the given reactions. Our target is \( 2\text{C}_{(s)} + 2\text{H}_{2(g)} \rightarrow \text{C}_2\text{H}_{4(g)} \). - We need 2 moles of C\(_{(graphite)}\) on the reactant side. So, we take reaction (ii) and multiply it by 2. \( 2\text{C}_{(s)} + 2\text{O}_{2(g)} \rightarrow 2\text{CO}_{2(g)} \), \( \Delta H = 2 \times (-394) = -788 \) kJ - We need 2 moles of H\(_{2(g)}\) on the reactant side. So, we take reaction (iii) and multiply it by 2. \( 2\text{H}_{2(g)} + \text{O}_{2(g)} \rightarrow 2\text{H}_2\text{O}_{(l)} \), \( \Delta H = 2 \times (-286) = -572 \) kJ - We need 1 mole of C\(_2\)H\(_{4(g)}\) on the product side. The given reaction (i) has it as a reactant. So, we must reverse reaction (i). When we reverse a reaction, we change the sign of its \( \Delta H \). \( 2\text{CO}_{2(g)} + 2\text{H}_2\text{O}_{(l)} \rightarrow \text{C}_2\text{H}_{4(g)} + 3\text{O}_{2(g)} \), \( \Delta H = -(-1411) = +1411 \) kJ Now, add the three manipulated equations together: (2 × ii): \( 2\text{C}_{(s)} + 2\text{O}_{2(g)} \rightarrow 2\text{CO}_{2(g)} \) (2 × iii): \( 2\text{H}_{2(g)} + \text{O}_{2(g)} \rightarrow 2\text{H}_2\text{O}_{(l)} \) (rev i): \( 2\text{CO}_{2(g)} + 2\text{H}_2\text{O}_{(l)} \rightarrow \text{C}_2\text{H}_{4(g)} + 3\text{O}_{2(g)} \) ------------------------------------------------------------------ Sum: \( 2\text{C}_{(s)} + 2\text{H}_{2(g)} + 3\text{O}_{2(g)} + 2\text{CO}_{2(g)} + 2\text{H}_2\text{O}_{(l)} \rightarrow 2\text{CO}_{2(g)} + 2\text{H}_2\text{O}_{(l)} + \text{C}_2\text{H}_{4(g)} + 3\text{O}_{2(g)} \) Cancel the species that appear on both sides (2CO\(_2\), 2H\(_2\)O, 3O\(_2\)): \[ 2\text{C}_{(s)} + 2\text{H}_{2(g)} \rightarrow \text{C}_2\text{H}_{4(g)} \] This is our target reaction. The enthalpy of formation is the sum of the enthalpy changes of the manipulated reactions: \[ \Delta H_f^\circ(\text{C}_2\text{H}_4) = (-788) + (-572) + (+1411) \] \[ \Delta H_f^\circ(\text{C}_2\text{H}_4) = -1360 + 1411 = +51 \text{ kJ/mol} \] Step 4: Final Answer:
The enthalpy of formation of C\(_2\)H\(_4\)\(_{(g)}\) is +51 kJ mol\(^{-1}\).