Question

Calculate change in enthalpy when 39 g acetylene is completely burnt with oxygen and enthalpy of combustion of acetylene is $-1300\ \text{kJ/mol}$. (At. mass C = 12, H = 1)

• A. $-975\ \text{kJ}$
• B. $-650\ \text{kJ}$
• C. $-1950\ \text{kJ}$
• D. $-1600\ \text{kJ}$

Hint:

1 mole of acetylene weighs 26 g. The problem asks for 39 g, which is exactly $1.5$ times the weight of 1 mole. Therefore, simply scale the energy by 1.5: $-1300 \times 1.5 = -1950\ \text{kJ}$!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem provides the molar enthalpy of combustion ($\Delta_c H^\circ$) for acetylene gas. We need to calculate the net heat released (enthalpy change) when a specific mass sample (39 g) of acetylene undergoes complete combustion.

Step 2: Key Formula or Approach:
The total enthalpy change ($\Delta H$) for a given mass is directly proportional to the number of moles consumed: $$ \Delta H = n \times \Delta_c H^\circ $$ where the number of moles ($n$) is: $$ n = \frac{\text{Given Mass}}{\text{Molar Mass}} $$

Step 3: Detailed Explanation:
First, let's compute the molar mass of acetylene ($\text{C}_2\text{H}_2$): $$ \text{Molar Mass} = (2 \times 12) + (2 \times 1) = 24 + 2 = 26\ \text{g/mol} $$ Next, calculate the total moles present in a $39\ \text{g}$ sample: $$ n = \frac{39\ \text{g}}{26\ \text{g/mol}} = 1.5\ \text{mol} $$ Now, multiply the number of moles by the standard molar enthalpy of combustion ($-1300\ \text{kJ/mol}$): $$ \Delta H = 1.5\ \text{mol} \times (-1300\ \text{kJ/mol}) = -1950\ \text{kJ} $$

Step 4: Final Answer: The net change in enthalpy during the combustion of 39 g of acetylene is $-1950\ \text{kJ}$, which corresponds to option (C).