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
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Ответ:
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.
Ответ:
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.
Ответ: