Question

The caloriefic value H₂(g) at STP is 12.78 KJ/L hence approximate standard enthalpy of formation of H₂O(l) is –

• A. –143 KJ
• B. –286 KJ
• C. Zero
• D. +286 KJ

Answer 1

Answer: (D)

+286 KJ

Answer 2

1 L H­₂(g) at STP = $\frac{1}{22.4}$ mol

\ Heat released due to combustion of $\frac{1}{22.4}$ mol of H₂(g)

= 12.78 KJ

Heat released due to combustion of 1 mol of H₂ (g)

= 12.78 × 22.4 = 286.27 KJ

\ approximate standard enthalpy of formation of H₂O(l) = –286 KJ