Question

The constant pressure molar heat capacities of linear gaseous molecules are approximately (7/2)R and those of non-linear gaseous molecules are approximately 4R. Estimate the change in the standard reaction entropy of the reaction forming liquid water from oxygen and hydrogen gas when the temperature is increased by 10K at constant pressure.

		-0.75 J/K
		-0.13 J/K
		+1.17 J/K
		+0.13 J/molK

Answer 1

The constant pressure molar heat capacities of linear gaseous molecules are approximately

(7/2)R and
those of non-linear gaseous molecules are approximately 4R.
Estimate the change in the standard reaction entropy of the reaction forming
liquid water from oxygen and hydrogen gas
when the temperature is increased by 10K at constant pressure.
      

Basis: 1 mole of O2
2H2(g) + O2(g) -> 2H2O(l)

2H2(g) + O2(g) -> 2H2O(g)
2H2O(g) -> 2H2O(l)

2H2(g) + O2(g) -> 2H2O(g)

H2 is a linear diatomic molecule, so Cp = 7/2*R
O2 is a linear diatomic molecule, so Cp = 7/2*R
H2O is a nonlinear polyatomic molecule, so Cp = 4*R

Let the reactants are initially at standard state, Ti = 298K
The final temperature, Tf = 298+10 = 308K

Change in entropy, DS = Int[dQ/T] = Int[(n*Cp*dT/T](Ti,Tf) = n*Cp*Ln(Tf/Ti)

Entropy change of H2, DS1 = 2*7*R/2*Ln(308/298)

Entropy change of O2, DS2 = 1*7*R/2*Ln(308/298)

Entropy change of H2O, DS3 = 2*4*R*Ln(308/298)

DSreaction = DSproduct - DSreactant = DS3 - DS1 - DS2 =
DSreaction = (2*4*R - 2*7*R/2 - 7*R/2) *Ln(308/298)= -2.5*R**Ln(308/298)
R = 8.314 J/mol-K
DSreaction = -2.5*8.314*Ln(308/298) = -0.686 J/mol-K

2H2O(g) -> 2H2O(l)
DSx = DHv/T
Enthalpy of vaporization, DHv = -40700 J/mol-K
DSx = -40700/308 = -132.14 J/mol-K

Total entropy change, DStotal = -0.686-132.14 = -132.83 J/mol-K