Question

A.) the standard enthalpy of formation of the reaction C2H4(g)+H2O(l)=C2H5OH(l)

B.) Use standard enthalpies of formation to calculate the standard enthalphy change of: the reaction of methane gas, CH4, with chlorine rine to form liquid chloroform, CHCl3. Gaseuous hydrogen chloridee is the other product.

C.) Use standard enthalpies of formation to calculate the standard enthalphy change of PCl3(g)+HCl(g)=PCl5(g)+H2(g)

Answer 1

A.

C2H4(g) + H2O(l) C2H5OH(l)

Standard enthalpies of formation:

H2O(l) = - 286kJ /mol;

C2H4(g) = 52.3 kJ/mol;

C2H5OH(l) = - 277 kJ/mol.

H_reaction = H_products - H_reactants

=(- 277 kJ/mol)- [ (52.3 kJ/mol) + (- 286 kJ /mol) ]

=(- 277 kJ/mol)- [ (52.3 kJ/mol) - (286 kJ /mol) ]

=(- 277 kJ/mol)- (- 233.7 kJ/mol)

=- 43.3 kJ/mol

B.

CH₄(g) + 3Cl₂(g) CHCl₃(l) + 3HCl(g)

Standard enthalpies of formation:

CHCl₃(l) = - 134.47 kJ /mol;

3HCl(g) = - 92.30 kJ/mol;

CH₄(g) = - 74.9 kJ/mol;

Cl₂(g) = 0 kJ/mol

H_reaction = H_products - H_reactants

=[ (- 134.47 kJ /mol) + 3 ( - 92.30 kJ/mol) ] - [ (- 74.9 kJ/mol) + 3 (0 kJ /mol) ]

= [ - 134.47 kJ /mol - 276.9 kJ/mol ] - [ (- 74.9 kJ/mol) ]

= [ - 411.37 kJ/mol + 74.9 kJ/mol ]

=- 336.47 kJ/mol

(C).

Note:

PCl₅(s) is solid not PCl₅(g)

PCl₃(g) + HCl(g) PCl₅(s) + H₂(g)

The Balanced equation is

PCl₃(g) + 2 HCl(g) PCl₅(s) + H₂(g)

Standard enthalpies of formation:

PCl₅(s) = - 440 kJ /mol;

H₂(g) = 0 kJ/mol;

PCl₃(g) = - 278 kJ/mol;

HCl(g) = - 92.30 kJ/mol

H_reaction = H_products - H_reactants

=[ (- 440 kJ /mol) + 2 ( 0 kJ/mol) ] - [ (- 278 kJ/mol) + 2 (- 92.30 kJ/mol) ]

=[ (- 440 kJ /mol) + 0 kJ/mol) ] - [ - 278 kJ/mol - 184.6 kJ/mol ]

=- 440 kJ /mol - ( - 462.6 kJ/mol )

=- 440 kJ /mol + 462.6 kJ/mol

= 22.6 kJ /mol