Chapter 7 and a bit of 19

Enthalpy and Thermochemical stoichiometry

100

When 1 mol of a fuel burns at a constant pressure, it produces 3452 kJ of heat and does 11kJ of work. What are ΔE and ΔH for the combustion of the fuel?

ΔH = q_p, ΔE = q + w

ΔE = -3452 - 11 = -3463 kJ

200

Determine whether each process is exothermic or endothermic and indicate the sign of ΔH

a) natural gas burning on a stove

b) Isopropyl alcohol evaporating from skin

c) water condensing from steam

a) exothermic, - ΔH

b) endothermic, +ΔH

c) exothermic, -ΔH

300

What mass of natural gas (CH₄) must burn to emit 325 kJ of heat?

CH_4(g) + 2O_2(g) -> 2 H₂O_(g) + CO_2(g), ΔH^o_rxn = -802.3 kJ

325 kJ * (1mol/802.3kJ) * (16.04g/1mol) = 6.50g of CH₄

400

Consider the thermochemical equation for the combustion of acetone (C₃H₆O), the main ingredient in nail polish remover:

C₃H₆O_(l) + 4O_2(g)2 -> 3CO_2(g) + 3H₂O_(g), ΔH^o_rxn = -1790kJ

If a bottle of nail polish remover contains 237mL of acetone, how much heat is associated with its complete combustion?. d_acetone = 0.788g/mL

237mL (0.788g/1mL)(1mol/58.08g)(1790kJ/1mol) = -5.76 x 10³kJ released

500

The propane fuel is used in gas barbeques burns according to the thermochemical equation:

C₃H_8(g) + 5 O_2(g) -> 3 CO_2(g) + 4 H₂O_(g), ΔH^o_rxn = -2044kJ

If a pork roast must absorb 1.6x10³ kJ to fully cook, and if only 10% of the heat produced by the barbeque is actually absorbed by the roast, what mass of CO₂ is emitted into the atmosphere during the grilling of the pork roast?

1.6 x 10³ kJ used (100kJ generated/10kJ used)(3 mol CO₂/2044 kJ)(44.01g CO₂/1mol) = 1033.5 g CO₂

Thermal energy transfer and calorimetry

100

A 32.5-g iron rod, initially at 22.7 ^oC, is submerged into an unknown mass of water at 63.2 ^oC, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 59.5 ^oC. What is the mass of the water?

C_Fe = 0.449 J/g * ^oC, C_H2O= 4.18 J/ g * ^oC

q_lost = -q_gained

q_Fe = -q_H2O

m_FeC_FeΔT = -(m_H2OC_H2OΔT)

rearrange to solve for mass of water, the answer is 34.7g

200

Two substances, A and B, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of substance A is 6.15g and T_i = 20.5 ^oC. The mass of substance B is 25.2 g and T_i = 52.7 ^oC. The final temperature of both substances at thermal equilibrium is 46.7 ^oC. If the specific heat capacity of substance B is 1.17 J/g * ^oC, what is the specific heat capacity of A?

q_lost = -q_gained

m_BC_BΔT = -(m_AC_AΔT)

rearrange to solve for C_A, the answer is 1.10 J/g C

300

Exactly 1.5g of fuel burns under conditions of constant pressure and then again under conditions of constant volume. In measurement A the reaction produces 25.9 kJ of heat, and in measurement B the reaction produces 23.3 kJ of heat. Which measurement (A or B) corresponds to conditions of constant pressure? Which one corresponds to conditions of constant volume? Explain.

ΔH_rxn = q_p, ΔE_rxn = q_v = ΔH - PΔV. Combustions always involve expansions, expansions do work and therefore have a negative value. Combustions are also exothermic and have a negative ΔH value. Therefore, ΔE_rxn is more negative than ΔH^o_rxn; so A (-25.9kJ) is a constant-volume process and B is a constant pressure process

400

Zinc metals react with hydrochloric acid according to the balanced equation:

Zn_(s) + 2 HCl_(aq) -> ZnCl_2(aq) + H_2(g)

When 0.103g of Zn is combined with enough HCl to make 50.0mL of solution in a coffee-cup calorimeter, all of the zinc reacts, raising the temperature of the solution from 22.5 ^oC to 23.7 ^oC. Find ΔH_rxn for this reaction as written. Use 1.0 g/mL for the density of the solution and 4.18 J/g*C as the specific heat capacity)

At constant pressure, ΔH = q_cal = q_rxn

q_cal = q_soln =mC_solnΔT = 250.8J

q_soln = -q_rxn = -250.8J

0.103 g Zn (1mol/65.38g Zn) = 0.00158 mol Zn

ΔH_rxn = q_rxn/mol Zn = -1.6 x 10⁵ J/mol

500

Mothballs are composed primarily of the hydrocarbon naphthalene (C₁₀H₈). When 0.820g of naphthalene burns in a bomb calorimeter, the temperature rises from 25.10 ^oC to 31.56 ^oC. Find ΔE_rxn for the combustion of naphthalene. The heat capacity of the bomb calorimeter is 5.11 kJ/^oC

q_cal = CΔT = 33.01 kJ

q_cal = -q_rxn = -33.01 kJ

0.820g naphthalene = 0.00639 mol naphthalene

ΔE_rxn = q_rxn/ mol naphthalene = -5.16 x 10³ kJ/mol

Hess's law and enthalpies of formation

100

Calculate ΔH_rxn for the reaction

Fe₂O_3(s) + 3 CO_(g) -> 2 Fe_(s) + 3 CO_2(g)

Use the following reactions and given ΔH's:

1) 2 Fe(s) + 3/2 O_2(g) -> Fe₂O_3(s), ΔH = -824.2 kJ

2) CO_(g) + 1/2 O_2(g) -> CO_2(g), ΔH = -282.7 kJ

Because reaction (1) has Fe₂O₃ as a product and the overall reaction has it as a reactant, we need to reverse reaction (1). The ΔH will change the sign

Because reaction (2) has 1 mole CO as a reactant and the overall reaction has 3 moles of CO as reactant, we need to multiply reaction (2)  and ΔH by 3.

Hess's law states that the ΔH of the net reaction is the sum of the ΔH of the steps, so +824.2kJ - 848.1kJ = -23.9 kJ = ΔH_rxn

200

Calculate ΔH_rxn for the reaction:

CH_4(g) + 4 Cl_2(g) -> CCl_4(g) + 4 HCl_(g)

Use the following reactions and given ΔH's:

(1) C_(s) + 2H_2(g) -> CH_4(g), ΔH = -74.6 kJ

(2) C_(s) + 2Cl_2(g) -> CCl_4(g), ΔH = -95.7 kJ

(3) H_2(g) + Cl_2(g) -> 2 HCl_(g), ΔH = -92.3 kJ

We need to reverse reaction (1) because we need CH₄ as a reactant and not a product, ΔH will change its sign.

Reaction (2) remains unchanged

Reaction (3) and ΔH needs to be multiplied by 2 to match moles of Cl and HCl.

Hess's Law states that the ΔH of the net reaction is the sum of the ΔH of the steps, so 74.6kJ - 95.7 kJ - 184.6 kJ = -205.7 kJ = ΔH_rxn

300

Hydrazine (N₂H₄) is a fuel used by some spacecraft. It is normally oxidized by N₂O₄ according to the equation:

N₂H_4(l) + N₂O_4(g) -> 2N₂O_(g) + 2H₂O_(g)

Calculate ΔH^o_rxn for this reaction using standard enthalpies of formation

N₂H_4(l) has 50.6 kJ/mol ΔH^o_f

N₂O_4(g) has 9.16 kJ/mol ΔH^o_f

N₂O_(g) has 81.6 kJ/mol ΔH^o_f

2H₂O_(g) has -241.8 kJ/mol ΔH^o_f

ΔH^o_rxn = ∑n_pΔH^o_f (products) - ∑n_RΔH^o_f (reactants)

plug and chug the numbers, ΔH^o_rxn = -380.2 kJ

400

During photosynthesis, plants use energy from sunlight to form glucose (C₆H₁₂O₆) and oxygen from carbon dioxide and water. Write a balanced equation for photosynthesis and calculate ΔH^o_rxn

6CO_2(g) + 6H₂O_(l) -> C₆H₁₂O_6(s) + 6O_2(g)

CO_2(g) has ΔH^o_f of -393.5 kJ/mol

H₂O_(l) has ΔH^o_f of -285.8 kJ/mol

C₆H₁₂O_6(s) has ΔH^o_f of -1273.3 kJ/mol

O_2(g) has ΔH^o_f of 0

ΔH^o_rxn = ∑n_pΔH^o_f (products) - ∑n_RΔH^o_f (reactants)

plug and chug the numbers, ΔH^o_rxn = +2802.5 kJ

500

The explosive nitroglycerin (C₃H₅N₃O₉) decomposes rapidly upon ignition or sudden impact according to the balanced equation:

4 C₃H₅N₃O_9(l) -> 12 CO_2(g) + 10 H₂O_(g) + 6 N_2(g) + O_2(g), ΔH^o_rxn = -5678 kJ

Calculate the standard enthalpy of formation ΔH^o_f for nitroglycerin 

CO_2(g) has ΔH^o_f of -393.5 kJ/mol

H₂O_(g) has ΔH^o_f of -241.8 kJ/mol

N_2(g) has ΔH^o_f of 0

O_2(g) has ΔH^o_f of 0

ΔH^o_rxn = ∑n_pΔH^o_f (products) - ∑n_RΔH^o_f (reactants)

solve for ΔH^o_f  in terms of variables, then plug in numbers, ΔH^o_f = -365.5 kJ/mol

Entropy, 2nd law of Thermodynamic, and whatever

100

Which of these processes is spontaneous?

a) the combustion of natural gas

b) the extraction of iron metal from iron ore

c) a hot drink cooling to room temperature

d) drawing heat energy from the ocean's surface to power a ship

a and c are spontaneous

200

Which of these processes are nonspontaneous? Are the nonspontaneous processes impossible?

a) A bike going up a hill

b) a meteor falling to Earth

c) obtaining hydrogen gas from liquid water

d) a ball rolling down a hill

 a and c are nonspontaneous. Nonspontaneous processes are not impossible, we just need to do some sort of work to make these processes happen.

300

Calculate the change in entropy that occurs in the system when 1.50 mol of isopropyl alcohol (C₃H₈O) melts at its melting point (-89.5 ^oC). ΔH^o_fus = 5.37 kJ/mol

1.5 mol (5.37kJ/1mol) = 8.055kJ = 8.055x10³J

T = 273.15 - 89.5 = 183.65K

ΔS = q_rev/T  = 8.055x10³J / 183.65 K = 43.9 J/K

400

Without doing any calculations, determine the sign of ΔS_sys for each chemical reaction

a) 2KClO_3(s) -> 2KCl_(s) + 3O_2(g)

b) CH₂=CH_2(g) + H_2(g) -> CH₃CH_3(g)

c) Na_(s) + 1/2 Cl_2(g) -> NaCl_(s)

d) N_2(g) + 3H_2(g) -> 2NH_3(g)

a) ΔS > 0 because we're generating gas

b) ΔS < 0 because 2 moles of gas are converted to 1 mole of gas

c) ΔS < 0 because we're converting gas into solid

d) ΔS < 0 because we're converting 4 moles of gas to 2 moles of gas

500

Without doing any calculations, determine the signs of ΔS_sys and ΔS_surr for each chemical reaction. In addition, predict under what temperatures (all T, low T, or high T), if any, the reaction is spontaneous

a) 2 N_2(g) + O_2(g) -> 2 N₂O_(g), ΔH^o_rxn = +163.2 kJ

b) 4 NH_3(g) + 5 O_2(g) -> 4 NO_(g) + 6 H₂O_(g), ΔH^o_rxn = -906 kJ

c) C₃H_8(g) + 5 O_2(g) -> 3 CO_(g) + 4 H₂O ΔH^o_rxn = -2044kJ

a) ΔS_sys < 0 because 3 moles of gas are converted to 2 moles of gas. Because ΔH > 0, ΔS_surr < 0 therefore ΔS_universe < 0, therefore reaction is nonspontaneous at all T

 b) ΔS_sys > 0 because 9 moles of gas are converted to 10 moles of gas. Because ΔH < 0, ΔS_surr > 0 therefore ΔS_universe > 0, therefore reaction is spontaneous at all T

ΔS_sys > 0 because 6 moles of gas are converted to 7 moles of gas. Because ΔH < 0, ΔS_surr > 0 therefore ΔS_universe > 0, therefore reaction is spontaneous at all T