# Quiz | Thermochemistry

1

Which statement correctly describes an endothermic chemical reaction?

The products have higher potential energy than the reactants, and ∆H > 0

The products have higher potential energy than the reactants, and ∆H < 0

The products have lower potential energy than the reactants, and ∆H < 0

The products have lower potential energy than the reactants, and ∆H > 0

In an endothermic chemical reaction, reactants absorbs heat energy
⇒ the products will have higher potential energy than reactants
⇒ Hproducts > Hreactants
⇒ ∆H = Hproducts – Hreactants > 0

2

What happens to the value of ∆H of a thermochemical reaction if the reaction is reversed?

∆H has the same numerical value, and the sign remains the same

∆H has the same numerical value, and the sign changes

∆H has a different numerical value, but the sign remains the same

∆H has different numerical value and sign

Hess’s Law: enthalpy changes for chemical equations are additive
⇒ ΔHrxn = ΔHrxn (forward) + ΔHrxn (reverse) = 0
⇒ ΔHrxn (reverse) = - ΔHrxn (forward)

3

∆H = 114 kJ for the reaction: 2 NO2 (g) → 2 NO(g) + O2(g). What is the ∆H for NO(g) + ½ O2(g) → NO2(g)?

- 57 kJ

+ 57 kJ

- 114 kJ

+ 114 kJ

Reaction 1:   2 NO2 (g) → 2 NO(g) + O2(g),   Reaction 2:   NO(g) + ½ O2(g) → NO2(g)
Reaction 2 is the reverse of reaction 1 with the reactants and the products divided by 2

Hess’s Law: enthalpy changes for chemical equations are additive
⇒ ΔHrxn (reverse) = - ΔHrxn (forward)
AND   ΔHrxn = n x ΔHrxn (A) when a chemical equation A is multiplied by a factor n
⇒ ΔHrxn (Reaction 2) = - $\frac{1}{2}$ ΔHrxn (Reaction 1) = - $\frac{1}{2}$ x 114 = - 57 kJ

4

Which of the following does not have a standard heat of formation equal to 0 at 298 K and 1.0 atm?

H (g)

Ne (g)

N2 (g)

C (graphite)

The standard enthalpy of formation of the most stable form of an element is 0.
The most stable element for hydrogen is H2, not H ⇒ ΔH0f of H2 = 0   but   ΔH0f of H  0

5

Knowing that the enthalpy of combustion of propane gas is – 2220.0 kJ.mol-1, which of the following statements is incorrect?

Propane releases heat as it burns

Combustion of propane is an exothermic process

ΔH0f (C3H8, g) = -2220.0 kJ.mol-1

The enthalpy change of the combustion of 2 moles is – 4440.0 kJ

ΔH combustion (C3H8) = – 2220.0 kJ.mol-1 < 0 ⇒ exothermic process
Exothermic process ⇒ q < 0: reaction releases energy as heat
ΔH combustion (2 moles of C3H8) = 2 x ΔH combustion (C3H8) = – 4440.0 kJ

6

If a reaction is carried out at constant pressure, which of the following statements is correct?

The reaction is likely to be exothermic

The reaction is likely to be endothermic

The heat change is equal to the change of temperature

The heat change is equal to the enthalpy change

ΔHrxn = ΔUrxn + PΔVrxn     (when P = cst, isobaric conditions)
AND   ΔUrxn =  q + w = q - PΔVrxn    with q = heat change
⇒ ΔHrxn = q - PΔVrxn + PΔVrxn = q     at constant pressure