Question

at 298 K and constant pressure the heat of formation of C2H2 and C2H6 is 230 and 85 kJpermole . what is the enthalpy change in reaction ? 3C2H2 (g) -C6H6 (g)​

Answer 1

Given info : at 298 K and constant pressure the heat of formation of C2H2 and C2H6 is 230 and 85 kJ per mole .

To find : the enthalpy change in the reaction, 3C2H2 (g) ⇒C6H6(g)

Solution : reaction is .. 3C2H2(g) ⇒C6H6(g)

Enthalpy change = heat of formation of products - heat of formation of reactants.

= 85 kJ/mole - 3 × 230 kJ/mole

= 85 kJ/mole - 690 kJ/mole

= -605 kJ/mole

Therefore the change in enthalpy in the reaction is given by, -605 kJ/mole.

Answer 2

reaction =3C2H2(g) ⇒C6H6(g)

Enthalpy change = heat of formation of products - heat of formation of reactants.

= 85 kJ/mole - 3 × 230 kJ/mole

= 85 kJ/mole - 690 kJ/mole

= -605 kJ/mole

• Therefore the change in enthalpy in the reaction is given by, -605 kJ/mole.