Find qtrans, qrot, qvib, and qelectric and U Energy for the reactants H2O and CO2 using the balanced equation: CH4+2O2->2 H2O +CO2 at T=9335K, V=0.0345m^3 with 1.08 mol methane and 1.62 mol oxygen.

H2O: N=1.3E+24, De=1.6E-18, symmetry number=2, qvib1=5.3E+03, qvib2=2.3E+03, qvib3=5.4E+04 qvib sum=3. Vib Energy ground state= 9E-20.

CO2: N= 6.5E+23, De=2.7E-18, symmetry number=2, qvib1=1.9E+03, qvib2=3.4E+03, qvib3=9.6E+04, qvib sum=3. Vib Energy g.s.=5.0E-20.

We can now use the results of Sections 18-7 and 18-8 to construct q (V, T) for polyatomic molecules. For an ideal gas of linear polyatomic molecules, q ( V, T) is the product of Equations 18.43, 18.44, 18.46, and 18.50: T 3n-5 e-vib; /2T q ( V, T ) = 2It MKBT 3 V h2 (18.57) rot i=1 -@vib,j / T The energy is U vib.; / T D NIW + NIN NKQT + (18.58) 2T vib./ T _ K T and the heat capacity is Cy NIN + "vib.j -evib.; / T 2 + -vib. ;/ 7 \\ 2 (18.59) NKB 1=1 (1 18-9. Calculated Molar Heat Capacities Are in Very Good Agreement with Experimental Data 755 For an ideal gas of nonlinear polyatomic molecules, 1/2 2n MKRT\\ 3, JT 1/2 T3 q(V, T) = V . J h2 0 rot,A rot, B, rot, C (18.60) 3n-6 e -vih,; /2T X (1 - e- vib; /T U 3nt-6 vib,j vib., / T + NIW + (18.61) NK T + E De NIW 1=1 27 vib. j / T K T and Cv 3n-6 e - vib, } / T NIW + vib.j NIW NKB + (18.62) (1 - e-vib, ;/ T )2