The standard heat of reaction for the gas-phase dehydrogenation of propane (C3H8) to produce propene (C3H6) and H2 C3H8 → C3H6 + H2 is +124 kJ/mole. What would you predict for the heat of reaction of the following gas-phase process (in kJ/mole)? 2 C3H6 + 2 H2 → 2 C3H

Explanation:

The standard heat of reaction is the heat change when all the substances involved in a reaction have changed completely. it is the heat content of the reaction i.e the amount of heat involved in a chemical reaction.

For the reaction given ; C3H8 → C3H6 + H2 ; heat of reaction = +124 kJ/mole

2 C3H6 + 2 H2 → 2 C3H8, as such the heat of reaction is of backward reaction which will be negative, hence the prediction of the heat of reaction for the gas phase reaction will be -248KJ/mole.

The answer is -248.0 kJ/Mole

Explanation:

The heat of reaction is the total amount of energy either absorbed or released during a chemical reaction.

Heat of reaction, ΔH° rxn = ∑ nΔH°f products - ∑nΔH°f reactants

From heat of formation tables (standard reference materials)

Heat of formation ΔH°f of C3H8 = -103.85 kJ/mole

Heat of formation for all pure elements = 0

where n = no. of moles

We have two reactions in this experiment: a forward and backward reactions with different number of moles of reactants and products.

C3H8 → C3H6 + H2     (ΔH° = +124kJ/mole)  (1)

2C3H6 + 2 H2 → 2C3H8   (ΔH° = ?? = x)(2)

From eqn. (1),

ΔH°rxn = (ΔH°f C3H6 + ΔH°f H2) - ( ΔH°f C3H8)

124 = ( ΔH°f C3H6 + 0) - (-103.85)

124 = ΔH°f C3H6 + 103.85

⇒ΔH°f C3H6  = 124- 103.85

= 20.15 kJ/mole.

∴ solving ΔH°f C3H6   into eqn (2)

ΔH° =x = (2 × (-103.85)) - (2 × 20.15 + 2(0))

x= -207.7 - 40.3

= -248.0

∴ ΔH° = -248.0 kJ/mole.