Calculate the enthalpy of vaporization of liquid water in kJ/mol using the following data: (i). H2(g)+((1/2))O2(g) arrow H2O(.l.) Δ H=-285.77 kJ/mol (ii). H2(g)+((1/2))O2(g) arrow H2O(.g.) , Δ H=-241.84 k J / m o l

Question

Calculate the enthalpy of vaporization of liquid water in kJ/mol using the following data: (i). H2(g)+((1/2))O2(g) arrow H2O(.l.) Δ H=-285.77 kJ/mol (ii). H2(g)+((1/2))O2(g) arrow H2O(.g.) , Δ H=-241.84 k J / m o l (A) +43.93 (B) -43.93 (C) +527.61 (D) -527.61

Tardigrade Admin · Accepted Answer

Enthalpy of vapourisation of textH text2 textO is H2O(l) arrow H2O(g) i.e. (ii)-(i) Enthalpy of vaporization of liquid water =285.77-241.84=+43.93 kJ mol- 1

Q. Calculate the enthalpy of vaporization of liquid water in $k J / m o l$ using the following data:
(i). $H_{2} (g) + (\frac{1}{2}) O_{2} (g) \to H_{2} O (l)$
$Δ H = - 285.77 k J / m o l$
(ii). $H_{2} (g) + (\frac{1}{2}) O_{2} (g) \to H_{2} O (g)$ ,
$Δ H = - 241.84 k J / m o l$

$+ 43.93$

$- 43.93$

$+ 527.61$

$- 527.61$

Enthalpy of vapourisation of $H_{2} O$ is
$H_{2} O (l) \to H_{2} O (g)$
i.e. (ii)-(i)
Enthalpy of vaporization of liquid water $= 285.77 - 241.84 = + 43.93 k J m o l^{- 1}$

Q. Calculate the enthalpy of vaporization of liquid water in kJ/mol using the following data: (i). H2​(g)+(21​)O2​(g)→H2​O(l) ΔH=−285.77kJ/mol (ii). H2​(g)+(21​)O2​(g)→H2​O(g) , ΔH=−241.84kJ/mol

+43.93

−43.93

+527.61

−527.61