Hess Law

Hess Law

By Angela Toh

Multiple Choice

Hess' Law makes use of which principle to calculate the enthalpy change of a reaction?

• The law of conservation of energy

• The law of conservation of matter

• The law that you will always find a lost item in the last place you look for it

• Murphy's law

Multiple Choice

What is the Delta H if you reverse the reaction.

• 86 kJ

• - 86 kJ

• 68 kJ

• -68 kJ

​Assignment 17.2 Question 1

Multiple Choice

The following equations show the oxidation of carbon and carbon monoxide to carbon dioxide.

C_(s) +O_2(g) → CO_2(g) ΔH = –x kJ mol^–1

CO_(g) + O_2(g) → CO_2(g) ΔH = –y kJ mol^–1

What is the enthalpy change, in kJ mol^–1, for the oxidation of carbon to carbon monoxide?

C_(s) + O_2(g) → CO_(g)

• x + y

• -x - y

• y - x

• x - y

Multiple Choice

Consider the following equations.

Mg_(s) + O_2(g) → MgO_(s) ∆H = –602 kJ

H_2(g) + O_2(g) → H₂O_(g) ∆H = –242 kJ

What is the ∆H value (in kJ) for the following reaction?

MgO_(s) + H_2(g) → Mg_(s) + H₂O_(g)

• -844

• -360

• +360

• +844

Multiple Choice

Using the equations below:

C_(s) + O_2(g) → CO_2(g) ∆H = –390 kJ

Mn_(s) + O_2(g) → MnO_2(s) ∆H = –520 kJ

what is ∆H (in kJ) for the following reaction?

MnO_2(s) + C_(s) → Mn_(s) + CO_2(g)

• 910

• 130

• -130

• -910

Multiple Choice

Using the equations below

Cu_(s) + ¹/₂O_2(g) → CuO_(s) ∆H = –156 kJ

2Cu_(s) + O_2(g) → Cu₂O_(s) ∆H = –170 kJ

what is the value of ∆H (in kJ) for the following reaction?

2CuO_(s) → Cu₂O_(s) + ¹/₂O_2(g)

• 142

• 15

• -15

• -142

Multiple Choice

The standard enthalpy change of formation values of two oxides of phosphorus are:

P_4(s) + 3O_2(g) → P₄O_6(s) ΔH_f = –1600 kJ mol^–1

P_4(s) + 5O_2(g) → P₄O_10(s) ΔH_f = –3000 kJ mol^–1

What is the enthalpy change, in kJ mol^–1, for the reaction below?

P₄O_6(s) + 2O_2(g) → P₄O_10(s)

• +4600

• +1400

• –1400

• –4600

Multiple Choice

The enthalpies of combustion of C(s), H2(g) and C4H9OH(l) (in kJmol^-1) are as follows   C(s) + O₂(g)   ->  CO₂(g)                                               ∆H=a H₂(g) + ½O₂(g)   ->   H₂O(l)                                        ∆H=b C₄H₉OH(l) + 6O₂(g)   ->   4CO₂(g) + 5H₂O(l)   ∆H=c

What is the enthalpy change for the reaction shown below?  

4C(g) + 5H₂(l) + ½O₂(g)   ->   C₄H₉OH(l)

• c – 4a – 5b
• 2a + 10b - c
• 4a + 5b - c
• 2a + 5b + c

​Assignment 17.2 Question 1