H 2(g)+ O 2(g) arrow H 2 O (l) ; Δ H298 K =-68.32 kcal. Heat of vaporization of water at 1 atm and 25° C is 10.52 kcal. The standard heat of formation (in kcal) of 1 mole of water vapour at 25° C is

Question

H 2(g)+ O 2(g) arrow H 2 O (l) ; Δ H298 K =-68.32 kcal. Heat of vaporization of water at 1 atm and 25° C is 10.52 kcal. The standard heat of formation (in kcal) of 1 mole of water vapour at 25° C is (A) -78.84 (B) 78.84 (C) +57.80 (D) -57.80

Tardigrade Admin · Accepted Answer

H 2(g)+(1/2) O 2(g) arrow H 2 O (l) ; Δ H=-68.32 kcal Vaporisation of water H 2 O (l) arrow H 2 O (g) ; Δ H=10.52 kcal Here Δ H=Δ Hf[ H 2 O (g)]-Δ Hf[ H 2 O (l)] Δ Hf[ H 2 O (g)]=10.52+(-68.32)=-57.80 kcal

Q. $H_{2 (g)} + O_{2 (g)} \to H_{2} O_{(l)}; Δ H_{298 K} = - 68.32 k c a l$ . Heat of vaporization of water at $1$ atm and $2 5^{\circ} C$ is $10.52 k c a l$ . The standard heat of formation (in kcal) of $1$ mole of water vapour at $2 5^{\circ} C$ is

-78.84

78.84

+57.80

-57.80

$H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(l)}; Δ H = - 68.32$ kcal

Vaporisation of water

$H_{2} O_{(l)} \to H_{2} O_{(g)}; Δ H = 10.52$ kcal

Here $Δ H = Δ H_{f [H_{2} O_{(g)}]} - Δ H_{f [H_{2} O_{(l)}]}$

$Δ H_{f} [H_{2} O_{(g)}] = 10.52 + (- 68.32) = - 57.80$ kcal

Q. H2(g)​+O2(g)​→H2​O(l)​;ΔH298K​=−68.32kcal. Heat of vaporization of water at 1 atm and 25∘C is 10.52kcal. The standard heat of formation (in kcal) of 1 mole of water vapour at 25∘C is