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1

Calcium reacts completely with excess water to produce calcium hydroxide and hydrogen.

Hydrogen gas and oxygen gas react according to the following equations:

(1.) How much more energy is stored in one mole of hydrogen molecules and 0.5 mole of oxygen molecules than is stored in one mole of gaseous water molecules?

(2.) What is the enthalpy change for the formation of one mole of liquid water from hydrogen gas and oxygen gas?

Suitable experiments have given the following results:

A reaction involved in the production of Fe from iron ore is

(1.) What mass of CO must react to release 54 kJ of heat?

(2.) What volume of CO at 298 K and 1.00 atmosphere is needed to produce 1.0 kg of Fe?

For the photosynthesis of fructose according to the equation

(1.) Calculate the standard heat of formation of fructose.

(2.) How many kilojoule of heat energy are required by a plant to produce 50.0 g of fructose?

Hydrogen gas (2.00 mole) reacts with excess chlorine and the heat liberated is used with 100% efficiency to convert a mixture of excess N₂ and excess O₂ into N₂O. What mass of N₂O is formed if all reactants and products are in the gaseous state?

At room temperature, nitrogen trichloride, NCl₃, is an unstable yellow oil. When exposed to strong light it explodes, producing only N₂ and Cl₂, and releasing 230 kilojoule of heat per mole of NCl₃.

(1.) Write an equation for the reaction.

(2.) Calculate ΔH° for the reaction as written.

(3.) Calculate the change in enthalpy accompanying the complete decomposition of a specimen of NCl₃ having a mass of 24.0 g.

(4.) If this specimen were completely decomposed, what volume of gas would be produced if measurements were made at 100 ° C and 4.00 atmosphere?

Using values given in the table of Average Bond Enthalpies, calculate the heats of hydrogenation of (1.) N₂ to the entirely single bonded product hydrazine, N₂H₄ (g), and (2.) C₂H₂ to the entirely singly bonded product ethane, C₂H₆ (g).

Normal octane, C₈H₁₈, has an unbranched chain of single bonded carbon atoms. The density of octane at 298 K is 0.704 g cm^-3 and H_v is 42 kJ mol^-1. Calculate from average bond enthalpies the heat evolved in the combustion of 4.00 litre of liquid octane in excess oxygen at 298 K and 1.00 atmosphere pressure.

Calculate the bond energy of the oxygen-oxygen bond in the hydrogen peroxide molecule given the following enthalpy changes:

1

Ca(s) + 2H₂O(l) → Ca(OH)₂(s) + H₂(g)

Sum of enthalpies of formation of products,

ΣΔH_f° (products) = ΔH_f°[Ca(OH)₂(s)] + ΔH_f°[H₂(g)]

= (-986 kJ mol^-1) + (0 kJ mol^-1) = -986 kJ mol^-1.

∴ ΔH° = ΣΔH_f° (products) - ΣΔH_f° (reactants)

= (-986 kJ mol^-1) - (-570 kJ mol^-1) = -416 kJ mol^-1.

This is for the reaction of 1 mol Ca.

The enthalpy change for 0.3 mol Ca = -416 kJ mol^-1 x 0.3 mol

= -125 kJ.

H₂(g) + ½O₂(g) → H₂O(g) (ΔH° = (-484 / 2) kJ mol^-1 = -242 kJ mol^-1

The negativeΔH° value indicates the reaction is exothermic, and represents the difference in energy contained in the reactant bonds (1 mol H₂(g) and 0.5 mol O₂(g)), and the energy in the product bonds (1 mol H₂O(g)). Therefore there is 242 kJ mol^-1 more energy stored in the reacting 1 mol H₂, 0.5 mol O₂, than in the 1 mol H₂O produced.

2H₂(g) + O₂(g) → 2H₂O(l) (ΔH° = -570 kJ mol^-1)

This represents the enthalpy change for the production of 2 mol H₂O(l).

The production of 1 mol H₂O(l) can be found by dividing the equation (and the enthalpy of reaction) by 2.

ThereforeΔH° for production of 1 mol H₂O(l) = (-570 / 2) kJ mol^-1

= -285 kJ mol^-1.

This question can be approached in two slightly different ways. The first uses Hess' law to combine all the given data into a final equation. The second method approaches the problem in smaller parts. Both answers are valid, but the second takes significantly longer.

First method:

Combine reaction equation 1 with reaction equation 3 and the reverse of reaction equation2. The overall reaction resuting is the desired equation, thus:

To calculate ΔH° for the final reaction requires the values of ΔH_f°[H₂O(g)] and ΔH_f°[PbO(s)]. Note that ΔH_f° for Pb(s) and H₂(g) are both zero as they are elements in their standard states.

The ΔH_f° value of PbO(s) is given by the second equation:

ΔH_f°[PbO(s)] = -219 kJ mol^-1

The value of ΔH_f°[H₂O(g)] can be found by combining the data given for the first and the third reaction equations:

From the first equation,ΔH_f°[H₂O(l)] = -285 kJ mol^-1

From the third equation,

H₂O(l) → H₂O(g) (ΔH° = +43 kJ mol^-1)

ΔH° = ΣΔH_f° (products) - ΣΔH_f° (reactants)

+43 kJ mol^-1 = ΔH_f°[H₂O(g)] - (-285 kJ mol^-1)

∴ ΔH_f°[H₂O(g)] = +43 kJ mol^-1 - 285 kJ mol^-1 = -242 kJ mol^-1

Now the final equation can be considered using the ΔH_f° values obtained:

PbO(s) + H₂(g) → Pb(s) + H₂O(g)

Sum of enthalpies of formation of reactants,

ΣΔH_f° (reactants) = ΔH_f°[PbO(s)] + ΔH_f°[H₂(g)]

= (-219 kJ mol^-1) + (0 kJ mol^-1) = -219 kJ mol^-1.

Sum of enthalpies of formation of products,

ΣΔH_f° (products) = ΔH_f°[Pb(s)] + ΔH_f°[H₂O(g)]

= (0 kJ mol^-1) + (-242 kJ mol^-1) = -242 kJ mol^-1.

∴ ΔH° = ΣΔH_f° (products) - ΣΔH_f° (reactants)

= (-242 kJ mol^-1) - (-219 kJ mol^-1) = -23 kJ mol^-1.

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g) (ΔH = -18 kJ mol^-1)

From the equation, 3.0 mol CO reacts to produce 18 kJ heat.

Therefore moles of CO required to produce 54 kJ heat

= 3.0 mol x (54 kJ / 18 kJ) = 3.0 mol x 3.0 = 9.0 mol.

Mass CO = moles of CO x molar mass CO

= 9.0 x 28.01 = 2.5 x 10² g.

From the balanced equation and given ΔH° value we can solve for ΔH_f°[C₆H₁₂O₆(s)].

Sum of enthalpies of formation of reactants,

ΣΔH_f° (reactants) = 6 x ΔH_f°[CO₂(g)] + 6 x ΔH_f°[H₂O(l)]

= (6 x -393 kJ mol^-1) + (6 x -285 kJ mol^-1) = -4068 kJ mol^-1.

Sum of enthalpies of formation of products,

ΣΔH_f° (products) = ΔH_f°[C₆H₁₂O₆(s)] + 6 x ΔH_f°[O₂(g)]

= ΔH_f°[C₆H₁₂O₆(s)] + (0 kJ mol^-1) = ΔH_f°[C₆H₁₂O₆(s)].

ΔH° = ΣΔH_f° (products) - ΣΔH_f° (reactants)

ΣΔH_f° (products) = ΔH° + ΣΔH_f° (reactants)

∴ ΔH_f°[C₆H₁₂O₆(s)] = 2828 kJ mol^-1 + (-4068 kJ mol^-1)

= -1240 kJ mol^-1.

Balanced equation:

H₂(g) + Cl₂(g) → 2HCl(g)

Sum of enthalpies of formation of reactants,

ΣΔH_f° (reactants) = ΔH_f°[H₂(g)] + ΔH_f°[Cl₂(g)]

= (0 kJ mol^-1) + (0 kJ mol^-1) = 0 kJ mol^-1.

Sum of enthalpies of formation of products,

ΣΔH_f° (products) = 2.00 x ΔH_f°[HCl(g)] = (2.00 x -92 kJ mol^-1)

= -184 kJ mol^-1.

∴ ΔH° = ΣΔH_f° (products) - ΣΔH_f° (reactants)

= (-184 kJ mol^-1) - (0 kJ mol^-1) = -184 kJ mol^-1.

Since chlorine is in excess in the first reaction, hydrogen is the limiting reactant and will determine the amount of heat produced:

Heat produced = no. mol H₂ x ΔH°

= 2.00 mol x -184 kJ mol^-1 = 368 kJ.

This energy will be the limiting factor in determining how much N₂ and O₂ will react.

ΔH° for the reaction N₂(g) + ½O₂(g) → N₂O(g) needs to be found:

Sum of enthalpies of formation of reactants,

ΣΔH_f° (reactants) = ΔH_f°[N₂(g)] + 0.5 x ΔH_f°[O₂(g)]

= (0 kJ mol^-1) + (0 kJ mol^-1) = 0 kJ mol^-1.

Sum of enthalpies of formation of products,

ΣΔH_f° (products) = ΔH_f°[N₂O(g)] = +82 kJ mol^-1.

∴ ΔH° = ΣΔH_f° (products) - ΣΔH_f° (reactants)

= (+82 kJ mol^-1) - (0 kJ mol^-1) = +82 kJ mol^-1.

Therefore each mole of N₂O requries 82 kJ energy (assuming 100% efficiency).

Moles of N₂O produced by 368 kJ = 368 kJ / 82 kJ mol^-1

= 4.49 mol.

Mass N₂O = no. mol N₂O x molar mass N₂O

= 4.49 mol x 44.02 g mol^-1 = 198 g.

Note that the question refers to the volume of the gases produced, and using the ideal gas equations the fact that the gas is a mixture of two different gases is not important - the only important factor is the number of moles present.

From the balanced equation, 1 mol NCl₃(l) gives 0.5 mol N₂(g) and 1.5 mol Cl₂(g), ie 2 moles of gaseous products formed per mol NCl₃.

∴ 0.200 mol NCl₃(l) produces 0.400 mol gas.

∴ V = (0.400 mol x 0.0821 L atm K^-1 mol^-1 x 373 K) / 4.00 atm = 3.06 L.

The heat of reaction is found from the relationship

ΔH_f° = ΔH°_{reactant bonds broken} + ΔH°_{product bonds formed}.

ΔH°_{reactant bonds broken} = ΔH°[N≡ N] + 2 x ΔH°[H-H]

= 945 kJ mol^-1 + (2 x 436 kJ mol^-1) = +1817 kJ mol^-1.

ΔH°_{product bonds formed} = ΔH°[N-N] + 4 x ΔH°[N-H]

= -163 kJ mol^-1 + (4 x -391 kJ mol^-1) = -1727 kJ mol^-1.

Overall reaction

ΔH° = ΔH°_{reactant bonds broken} + ΔH°_{product bonds formed}

= +1817 kJ mol^-1 + (-1727 kJ mol^-1)

= +90 kJ mol^-1.

(2.) For the reaction C₂H₄(g) + H₂(g) → C₂H₆(g):

Bonds broken (ΔH positive)

ΔH°_{reactant bonds broken} = ΔH°[C=C] + 4 x ΔH°[C-H] + ΔH°[H-H]

= 614 kJ mol^-1 + (4 x 413 kJ mol^-1) + 436 kJ mol^-1

= 2702 kJ mol^-1.

ΔH°_{product bonds formed} = ΔH°[C-C] + 6 x ΔH°[C-H]

= -348 kJ mol^-1 + (6 x -413 kJ mol^-1) = -2826 kJ mol^-1.

Overall reaction

ΔH° = ΔH°_{reactant bonds broken} + ΔH°_{product bonds formed}

= 2702 kJ mol^-1 + (-2826 kJ mol^-1) = -124 kJ mol^-1.

For the reaction C₈H₁₈(g) + 12½O₂(g) → 8CO₂(g) + 9H₂O(g)

ΔH°_{reactant bonds broken} = 7 x ΔH° [C-C] + 18 x ΔH° [C-H] + 12.5 x ΔH° [O=O] = (7 x 348 kJ mol^-1) + (18 x 413 kJ mol^-1) + (12.5 x 498 kJ mol^-1) = 16095 kJ mol^-1.

ΔH°_{product bonds formed} = 16 x ΔH° [C=O] + 18 x ΔH⁰[O-H] = (16 x -745 kJ mol^-1) + (18 x -463 kJ mol^-1) = -20254 kJ mol^-1.

∴ ΔH° = ΔH°_{reactant bonds broken} + ΔH°_{product bonds formed} = 16095 kJ mol^-1 + (-20254 kJ mol^-1) = -4159 kJ mol^-1.

This ΔH° value needs to be corrected by the ΔH_v° value provided for octane, which is added to allow for the energy absorbed by the liquid octane to allow it to vaporise and burn, hence reducing the amount of heat released by the reaction:

Corrected ΔH° = -4159 kJ mol^-1 + 42 kJ mol^-1 = -4117 kJ mol^-1.

Mass C₈H₁₈ = Volume C₈H₁₈ x density C₈H₁₈ = 4.00 x 10³ cm³ x 0.704 g L^-1 = 2816 g. No. mol C₈H₁₈ = mass C₈H₁₈ / molar mass C₈H₁₈ = 2816 g / 114.22 g mol^-1 = 24.7 mol.

The heat released by the reaction was found to be -4117 kJ mol^-1, so the heat released by 4.00 L C₈H₁₈ = no. mol C₈H₁₈ x heat released per mole = 24.7 mol x -4117 kJ mol^-1 = 1.02 x 10⁵ kJ mol^-1.

For the reaction H₂(g) + O₂(g) → H₂O₂(g) (ΔH° = -136 kJ mol^-1)

ΔH°_{reactant bonds broken} = ΔH°[H₂(g) → 2H(g)] + ΔH°[O₂(g) → 2O(g)] = +436 kJ mol^-1 + 498 kJ mol^-1) = +934 kJ mol^-1.

ΔH°_{product bonds formed} = ΔH°[O-O] + 2 x ΔH°[O-H] - Equation A

Note that both ΔH°[O-O] and ΔH°[O-H] are unknown. In order to solve for ΔH°[O-O], we need to obtain ΔH°[O-H] from equation 2 in the question:

For the reaction H₂(g) + ½O₂(g) → H₂O(g) (ΔH = -242 kJ mol^-1)

ΔH°_{reactant bonds broken} = ΔH°[H₂(g) → 2H(g)] + 0.5 x ΔH°[O₂(g) → 2O(g)] = +436 kJ mol^-1 + (0.5 x 498 kJ mol^-1) = +685 kJ mol^-1.

ΔH°_{product bonds formed} = 2 x ΔH°[O-H]

ΔH° = -242 kJ mol^-1

Rearranging ΔH° = ΔH°_{reactant bonds broken} + ΔH°_{product bonds formed}:

ΔH°_{product bonds formed} = ΔH° - ΔH°_{reactant bonds broken}

∴ 2 x ΔH°[O-H] = -242 kJ mol^-1 - (+685 kJ mol^-1) = 927 kJ mol^-1.

∴ ΔH°[O-H] = 927 kJ mol^-1 / 2 = 464 kJ mol^-1.

Now that we have a value for ΔH°[O-H], we can return to equation A and solve for ΔH°[O-O]:

ΔH°_{product bonds formed} = ΔH°[O-O] + 2 x ΔH°[O-H] - Equation A

ΔH° = -136 kJ mol^-1

Rearranging ΔH° = ΔH°_{reactant bonds broken} + ΔH°_{product bonds formed}:

ΔH°_{product bonds formed} = ΔH° - ΔH°_{reactant bonds broken}

∴ ΔH°[O-O] + (2 x 464 kJ mol^-1) = -136 kJ mol^-1 + 934 kJ mol^-1

ΔH°[O-O] + 928 kJ mol^-1 = 798 kJ mol^-1

ΔH°[O-O] = -130 kJ mol^-1