In order to decompose 9 g of water 142.5 kJ of heat is required. Hence enthalpy of formation of water is

Question

In order to decompose 9 g of water 142.5 kJ of heat is required. Hence enthalpy of formation of water is (A) -142.5 kJ (B) +142.5 kJ (C) -285 kJ (D) 285 kJ

Tardigrade Admin · Accepted Answer

Heat required to decompose 1 mol (18 g) of H2O = 2 × 142.5 = 285 kJ Δ Hf is the reverse of heat required = 285 kJ mol-1

Q. In order to decompose 9 g of water 142.5 kJ of heat is required. Hence enthalpy of formation of water is

-142.5 kJ

+142.5 kJ

-285 kJ

285 kJ

Heat required to decompose 1 mol (18 g) of $H_{2} O$
= $2 \times 142.5$ = 285 kJ
$Δ H_{f}$ is the reverse of heat required
= 285 kJ $m o l^{- 1}$

Q. In order to decompose 9 g of water 142.5 kJ of heat is required. Hence enthalpy of formation of water is

-142.5 kJ

+142.5 kJ

-285 kJ

285 kJ

Heat required to decompose 1 mol (18 g) of H2​O = 2×142.5 = 285 kJ ΔHf​ is the reverse of heat required = 285 kJ mol−1

Heat required to decompose 1 mol (18 g) of $H_{2} O$
= $2 \times 142.5$ = 285 kJ
$Δ H_{f}$ is the reverse of heat required
= 285 kJ $m o l^{- 1}$