1. At what standard conditions does the heat of combustion calculated?
a) 26°C, 1 atm
b) 25°C, 0.5 atm
c) 25°C, 1 atm
d) 26°C, 0.5 atm
Explanation: The standard heat of combustion Δh°c is the specific enthalpy change associated with this reaction at standard conditions, usually 25°C and 1 atm pressure. The standard heat of reaction is the difference between the heats of combustion of reactants and products.
2. Hydrogen peroxide decomposes according to the following thermochemical reaction:
H2O2 (l) → H2O (l) + 1/2 O2(g); ΔH = -98.2 kJ
Calculate the change in enthalpy, ΔH, when 1.00 g of hydrogen peroxide decomposes.
a) -2.89 kJ
b) -2.80 kJ
c) -2.00 kJ
d) -2.85 kJ
Explanation: Molecular mass of H2O2 is 34.0 (2 x 1 for hydrogen + 2 x 16 for oxygen), which means that 1 mol H2O2 = 34.0 g H2O2.
Using these values:
ΔH = 1.00 g H2O2 x 1 mol H2O2 / 34.0 g H2O2 x -98.2 kJ / 1 mol H2O2
ΔH = -2.89 kJ.
3. Calculate ΔH if a piece of metal with a specific heat of 0.98 kJ-kg-1-K-1 and a mass of 2kg is heated from 22°C to 28°C.
a) 11.66 kg
b) 11.56 kg
c) 11.76 kg
d) 11.26 kg
Explanation: ΔH = q = cpsp × m × (ΔT) = (0.98) × (2) × (+6°) = 11.76 kg.
4. If a calorimeter’s ΔH is +2001 Joules, how much heat did the substance inside the cup lose?
a) +2000 J
b) – 2001 J
c) – 2000J
d) + 2001 J
Explanation: Since the heat gained by the calorimeter is equal to the heat lost by the system, then the substance inside must have lost the negative of +2001 J, which is – 2001 J.
5. Calculate the ΔH of the following reaction: CO2(g) + H2(g)O -> H2CO3(g) if the standard values of ΔHf are as follows: CO2(g) : -393.509 KJ/mol, H2O(g): – 241.83 KJ/mol, and H2CO3(g) : – 275.2 KJ/mol
a) +360.139 KJ
b) +350.129 KJ
c) – 360.139 KJ
d) -350.129 KJ
Explanation: ΔH° = ƩΔvpΔH°f (products) – ƩΔvrΔH°f (reactants) so this means that you add up the sum of the ΔH’s of the products and subtract away the ΔH of the products:
(- 275.2 KJ) – (-393.509 KJ + -241.83KJ) = (-275.2) – (-635.339) = + 360.139 KJ.
6. Calculate ΔH if a piece of aluminum with a specific heat of 0.9 kJ-kg-1-K-1 and a mass of 1.6 kg is heated from 286°K to 299°K.
a) 16.52 kJ
b) 17.72 kJ
c) 18.72 kJ
d) 15.52 kJ
Explanation: ΔH = q = cpsp × m × (ΔT) = (0.9) × (1.6) × (13) = 18.72 kJ.
7. Calculate the ΔH value of the reaction:
HCl + NH3 → NH4Cl
(ΔH values for HCl is -92.30; NH2 is -80.29; NH4Cl is -314.4).
a) 141.8
b) 121.8
c) 131.8
d) 151.8
Explanation: ΔH = ΔHproducts – ΔHreactants
ΔHproducts = -314.4
ΔHreactants = -92.30 + (-80.29) = -172.59
ΔH = -314.4 – 172.59 = 141.8.
8. Calculate ΔH for the reaction:
N2 + 3H2 → 2NH3
(The bond dissociation energy for N-N is 163 kJ/mol; H-H is 436 kJ/mol; N-H is 391 kJ/mol).
a) 867
b) 895
c) 847
d) 875
Explanation: ΔH = ΔHproducts – ΔHreactants
To use the bond dissociation energies, we must determine how many bonds are in the products and the reactants. In NH3 there are 3 N-H bonds so in 2 NH3 there are 6 N-H bonds. In N2 there is 1 N-N bond and in 3H2 there are 3 H-H bonds.
ΔHproducts = 6(391) = 2346
ΔHreactants = 163 + 3(436) = 1471
ΔH = 2346 – 1471 = 875.
9. Consider the reaction:
N2 + O2 → 2NO; ΔH = +180.6 kJ
what is the enthalpy change for the formation of one mole of nitrogen (II) oxide?
a) 96.80 kJ
b) 95.30 kJ
c) 96.50 kJ
d) 95.50 kJ
Explanation: Here we use the conversion factor 180.6 kJ/2 mol NO.
ΔH = 1 mol NO × (180.6 kJ/2 mol NO) = 95.30 kJ.
10. Refer to Q9 and answer what is the enthalpy change for the reaction of 100.0 g of nitrogen with excess oxygen?
a) 1164 kJ
b) 1190 kJ
c) 1194 kJ
d) 1160 kJ
Explanation: Here, we need to convert grams of N2 to moles of N2 and use the conversion factor 180.6 kJ/1 mol N2.
180.6 kJ × (1 mol N2/28.01 g N2) × (180.6 kJ/1 mol N2) = 1164 kJ.