Question

The standard enthalpy of formation of C\(_2\)H\(_4\)(g), CO\(_2\)(g), and H\(_2\)O(l) are 52, –394, and –286 kJ mol\(^{-1}\) respectively. The heat evolved by burning 7 g of C\(_2\)H\(_4\)(g) is

• A. 353 kJ
• B. 1412 kJ
• C. 706 kJ
• D. 1306 kJ

Hint:

Use the enthalpy of formation values to compute the enthalpy change of reactions. Always multiply by the stoichiometric coefficients from the balanced chemical equation.

Answer 1

The Correct Option is C

Step 1: Write the balanced combustion reaction.
\[ \text{C}_2\text{H}_4(g) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 2\text{H}_2\text{O}(l) \] Step 2: Use enthalpies of formation to calculate \(\Delta H\)
\[ \Delta H = \left[2(-394) + 2(-286)\right] - [1(52)]
= (-788 - 572) - 52 = -1360 - 52 = -1412 \text{ kJ/mol} \] Step 3: Find moles of C\(_2\)H\(_4\) in 7 g
\[ \text{Molar mass of C}_2\text{H}_4 = 2 \times 12 + 4 \times 1 = 28 \text{ g/mol}
\text{Moles} = \frac{7}{28} = 0.25 \] Step 4: Multiply with enthalpy change per mole
\[ \text{Heat evolved} = 0.25 \times 1412 = 353 \times 2 = 706 \text{ kJ} \]