Consider the following reaction at 257oC and 1.00 atm:
CH3Cl2(g) +H2(g) -------> CH4(g) + HCl(g)
for this reaction, the enthalpy change at 257oC is -83.4 kJ/mol.
You may need to use the following data:
CH3Cl(g) Cp 48.5 J x oC x mole
H2 (g) AH 0 Cp 28.9 J x oC x mole
CH4 (g) AH - 75 Cp 41.3 J x oC x mole
HCl (g) AH - 92 Cp 29.1 J x oC x mole
a) Assuming that the Cp values are independent, calculate AH for this reaction at 25oC.
b) Calculate AHf for CH3Cl