The standard enthalpy of formation of gaseous H2O at text298 textK is - text241 text.82 textkJ textmo textl- 1 . Calculate textΔH textf texto at text373 textK, given the following values of the molar heat capacities at constant pressure. Molar heat capacity of ( textH)2 textO ( textg) = 33.58 ( textJK)- 1 ( textmol)- 1 Molar heat capacity of H2(g)=28.84 text J( textK)- 1mol- 1 Molar heat capacity of O2(g)=29.37 text J( textK)- 1mol- 1 Assume that the heat capacities are independent of temperature.

Question

The standard enthalpy of formation of gaseous H2O at text298 textK is - text241 text.82 textkJ textmo textl- 1 . Calculate textΔH textf texto at text373 textK, given the following values of the molar heat capacities at constant pressure. Molar heat capacity of ( textH)2 textO ( textg) = 33.58 ( textJK)- 1 ( textmol)- 1 Molar heat capacity of H2(g)=28.84 text J( textK)- 1mol- 1 Molar heat capacity of O2(g)=29.37 text J( textK)- 1mol- 1 Assume that the heat capacities are independent of temperature. (A) -242.6 text kJ mol- 1 (B) -485.2 text kJ mol- 1 (C) -121.3 text kJ mol- 1 (D) -286.4 text kJ mol- 1

Tardigrade Admin · Accepted Answer

The reaction is H2(g)+(1/2)O2(g) arrow H2O(g) Δ Cpo=Cp , mo(H2 O , g)- Cp , mo (H2 , g) + (1/2) Cp , mo (O2) =33.58- 28.84 + (1/2) (29.37) =-9.94 text JK- 1mol- 1 Using Kirchhoff's equation, Δ H° (373 K)=Δ H° (298 K)+(T2 - T1)Δ Cpo =-241.82+(373 - 298)× (- 9.94) =-242.6 text kJ mo textl- 1

$- 242.6 kJ m o l^{- 1}$

$- 485.2 kJ m o l^{- 1}$

$- 121.3 kJ m o l^{- 1}$

$- 286.4 kJ m o l^{- 1}$

−242.6 kJ mol−1

−485.2 kJ mol−1

−121.3 kJ mol−1

−286.4 kJ mol−1

$- 242.6 kJ m o l^{- 1}$

$- 485.2 kJ m o l^{- 1}$

$- 121.3 kJ m o l^{- 1}$

$- 286.4 kJ m o l^{- 1}$