Internal energy change

Problem 2

Hydrogen chloride gas readily dissolves in water, releasing 75.3 kJ/mol of heat in the process. If one mole of HCl at 298 K and 1 atm pressure occupies 24.5 liters, find ΔU for the system when one mole of HCl dissolves in water under these conditions.

Solution: In this process the volume of liquid remains practically unchanged, so ΔV = –24.5 L.

The work done is w = –PΔV = –(1 atm)(–24.5 L) = 24.6 L-atm

(The work is positive because it is being done on the system as its volume decreases due to the dissolution of the gas into the much smaller volume of the solution.) Using the conversion factor 1 L-atm = 101.33 J mol^–1 and substituting in Eq. 3 (above) we obtain

ΔU= q +PΔV = –(75300 J) + [101.33 J/L-atm) × (24.5 L-atm)] = –72.82 kJ

In other words, if the gaseous HCl could dissolve without volume change, the heat released by the process (75.3 kJ) would cause the system’s internal energy to diminish by 75.3 kJ. But the disappearance of the gaseous phase reduces the volume of the system. This is equivalent to compression of the system by the pressure of the atmosphere performing work on it and consuming part of the energy that would otherwise be liberated, reducing the net value of ΔU to –72.82 kJ.

Thrmodynamic work

Problem 1

Find the amount of work done on the surroundings when 1 liter of an ideal gas, initially at a pressure of 10 atm, is allowed to expand at constant temperature to 10 liters by
a) reducing the external pressure to 1 atm in a single step,
b) reducing P first to 5 atm, and then to 1 atm,
c) allowing the gas to expand into an evacuated space so its total volume is 10 liters.

Solution. First, note that ΔV, which is a state function, is the same for each path:

V₂ = (10/1) × (1 L) = 10 L, so ΔV = 9 L.

For path (a), w = –(1 atm)× (9 L) = –9 L-atm.

For path (b), the work is calculated for each stage separately:

w = –(5 atm) × (2–1 L) – (1 atm) × (10–2 L) = –13 L-atm

For path (c) the process would be carried out by removing all weights from the piston in Fig. 1 so that the gas expands to 10 L against zero external pressure. In this case w = (0 atm) × 9 L = 0; that is, no work is done because there is no force to oppose the expansion.

1) Which of the following salts (high spin for paramagnetic samples) will have the strongest attraction to a magnet?

A. MnSO₄ B.CoSO₄ C.ZnSO₄ D.CaSO₄.

2) If V(CO)₆ is an 18-electron complex, what is the charge on the complex, if any?

A. +2; B.+1; C.0; D.-1

3) In a metal carbonyl complex, as more electron density moves from the metal d-orbitals to the CO p* orbital (d --> p*), what happens to the CO bond order?

A. it increases B.it decreases C.it stays the same.

4) Photo-excitation promotes an electron from the HOMO (highest occupied molecular orbital) to the LUMO (lowest unoccupied molecular orbital) of a molecule.

i.Which is the stronger oxidant?

A.ground state - GS; B.excited state - ES C.GS is as oxidant as the excited one

iiWhich is the stronger reductant?

A.ground state - GS B.excited state - ES C.GS is as reductant as the excited one

5) Although pure diamonds are colorless, some diamonds have color because of dopants. The Hope diamond, for example, is blue due to nitrogen (N) dopants. Based on its position in the periodic table, N in diamond acts as a(n):

A.neither donor nor acceptor B.donor C.acceptor

6.For the reaction [Fe₂O₃ (s) + 3/2 C (s) –> 2 Fe (s) + 3/2 CO₂ (g)] DH at 300 K is 233.8 kJ while DS is 279.16 J´K^–1. Assuming no change in either quantity with temperature, the temperature at which this reaction becomes spontaneous is about

A. 1 K B. 800 K C.1000 K D.1200 K

7. The rate of chemical reaction depends on the total number of collision between reactant molecules. The total number of collision is favored by

(A) smaller collision cross section and low molecular velocity;

(B) smaller collision cross section and high molecular velocity;

(C) larger collision cross section and low molecular velocity;

(D) larger collision cross section and higher molecular velocity.

8) In heterogeneous catalysis of reactions on solid surfaces, the effectiveness of a given catalyst is not determined by

(A) bulk density of the catalyst;

(B) surface roughness of the catalyst;

(C) number of active sites on the catalyst surface;

(D) degree of adsorption of the reactant on the catalyst surface.

Thermodynamics Test.

1. Calculate DU for a system that does 300 kJ of work on the surroundings when 150 kJ of heat are absorbed by the system.

A) +450 kJ B) 0 kJ C) -450 kJ D) -150 kJ E)+150 kJ

2. Calculate DU for a system that absorbs 325 kJ of heat and has 65 kJ of work done on the system.

A) -260 kJ B) +390 kJ C) 0 kJ D) -390 kJ E) +260 kJ

3. Calculate DU for a system that loses 225 kJ of heat and has 150 kJ of work done on the sysyem.

A) +375 kJ B) -375 kJ C) +75 kJ D) 0 kJ E) -75 kJ

4. Calculate DU for a system that loses 325 kJ of heat while doing 200 kJ of work on the surroundings.

A) +125 kJ B) -525 kJ C) 0 kJ D) -125 kJ E) +525 kJ

5. In a certain exothermic reaction at constant pressure, DH = -75 kJ and 35 kJ of work was required to make room for products. What is DU?

A) -40 kJ B) 0 kJ C) -110 kJ D) +110 kJ E) +40 kJ

6. In a certain endothermic reaction at constant pressure, DH = +175 kJ and 45 kJ of work was required to make room for products. What is DU?

A) +220 kJ B) -220 kJ C) -130 kJ D) 0 kJ E) +130 kJ

7. For a certain reaction at constant pressure, DU = -125 kJ and 22 kJ of expansion work is done by the system. What is DH for this process?

A) -125 kJ B) +147 kJ C) +103 kJ D) -147 kJ E) -103 kJ

8. For a certain reaction at constant pressure, DU = +45 kJ and 14 kJ of expansion work is done by the system. What is DH for this process?

A) -59 kJ B) +59 kJ C) +31 kJ D) +45 kJ E) -31 kJ

9. Calculate the work needed to make room for products in the combustion of 1 mole of CH4(g) to carbon dioxide and water vapor at STP (1 L * atm = 101 J).

A) -4.52 kJ B) -2.26 kJ C) -6.79 kJ D) -11.3 kJ E) no work is needed

10. Calculate the work needed to make room for products in the combustion of 1 mole of C3H8(g) to carbon dioxide and water vapor where reactants and products are brought to STP (1 L * atm = 101 J).

A) no work is needed B) -15.8 kJ C) -13.6 kJ

D) -4.52 kJ E) -2.26 kJ

Here D represent the delta.

Answer Thermodynamics

1. D 2. B 3. E 4. B 5. C

6. E 7. E 8. B 9. E 10. E

iit chemistry

II and III, which molecule has the larger dipole moment?

ans both equally.(why?)

Chemistry Problems

RH₂ (ion exchange resin) can replace Ca²⁺ in hard water as:

RH+ Ca²⁺ ® RCa + 2H⁺

1 liter of hard water passing through RH₂ has pH 2. Hence hardness in ppm of Ca²⁺ is:

(a) 200 (b) 100 (c) 50 (d) 125 Ans-a.

A solution contains Na₂CO₃ and NaHCO₃. 10 ml of the solution required 2.5 ml of 0.1 M H₂SO₄ for neutralization using phenolphthalein as indicator. Methyl orange is then added when a further 2.5 ml of 0.2 M H₂SO₄ was required. The amount of Na₂CO₃ and NaHCO₃ in 1 liter of the solution is:

(a) 5.3 g and 4.2 g (b) 3.3 g and 6.2 g (c) 4.2 g and 5.3 g (d) 6.2 g and 3.3 g

Ans----(a)

Problem based on Stoihiometry

A sample of polystyrene prepared by heating styrene with tribromobenzoyl peroxide in the absence of air has the formula Br_3, C₆H₃(C₈H₈)n. The number n varies with the condition of preparation. One sample of polystyrene prepared in this manner was found to contain 10.46 % bromine. What is the value of n? Ans.. n = 19