1.31 Activity

1. A system does 3 J of work on the surroundings, and 12 J of work are added to the system.  What is the energy change of the system? 9 J What is the energy change of the surroundings? -9 J

2. A quantity of 0.850 mol of an ideal gas initially at a pressure of 15.0 atm and 300K is allowed to expand isothermally until its final pressure is 1.00 atm.  Calculate the value of the work done if the expansion is carried out (a) against a vaccum, (b) against a constant external pressure of 1.00 atm, and (c) reversibly – use the equation w_rev=-nRT*ln(P₁/P₂).

a.) w=-P_extDV, here P_ext = 0, so w=0

b.) P_ext=1.00 atm.  Find initial and final volumes using the ideal gas law:

      V₁=nRT/P₁; V₂=nRT/P₂; w=-P_extDV, where P_ext is equal to P₂

      DV= V₂-V₁ = nRT({1/P₂}-{1/P₁})

      Therefore, w=-nRTP₂({1/P₂}-{1/P₁}) = -19.5 L*atm = -1.98*10³ J

c.) for an isothermal, reversible process, we use the following equation for work:

     w_rev=-nRT*ln(V₂/V₁)=-nRT*ln(P₂/P₁) = -5.74*10³ J

3. Calculate the values of DE and DH for the heating of 55.40 g of xenon from 300 K to 400 K.  Assume ideal-gas behavior and that the heat capacities at constant volume and constant pressure are independent of temperature.

Xenon is a monatomic “ideal” gas, so C_v=3/2*R and C_p=5/2*R (always true for ideal monoatomic gases).  We have the two following equations (always true):
DE=nC_vDT = 526J; DH=nC_pDT = 877 J

4. Calculate the entropy change when 2.0 mol of an ideal gas are allowed to expand isothermally from an initial volume of 1.5 L to 2.4 L.

DS=nR*ln(V₂/V₁) = 7.8 J/K

5. A quantity of 0.50 mol of an ideal gas at 20^oC expands isothermally against a pressure of 2.0 atm from 1.0 L to 5.0 L.  Calculate DS_sys, DS_surr, and DS_univ.

Use the ideal gas law to find initial pressure = 12 atm.  Use the equation:

DS_sys = nR*ln(V₂/V₁) = 6.7 J/K

DS_surr = q_surr/T; DE=0 since DT=0 (it is an isothermal process).  Therefore, w=-q.  We can calculate w=-PDV, and q_sys=-q_surr, so q_surr=w.  We get DS_surr = -2.8 J/K
DS_univ=DS_sys+DS_surr=3.9J/K

6. Supercooled water is liquid water that has been cooled below its normal freezing point.  This state is thermodynamically unstable and tends to freeze into ice spontaneously.  Suppose we have 2.0 mol of supercooled water turning to ice at -10^oC and 1.0 atm.  Calculate DS_sys, DS_surr, and DS_univ for this process (set up a thermodynamic cycle).  C_p water = 75.3 J/K*mol and C_p ice = 37.7 J/K*mol; the molar heat of fusion of water is 6.01 kJ/mol. (DS_fus=n*DH_fus/T, q_surr=n*DH_fus)

Step 1: Water at -10 à Water at 0:

DS = nC_p*ln(T_f/T_i) = 5.6 J/K

Step 2: Water freezes at 0:

DS_fus=DH_fus/T = -44.0 J/K

Step 3: Ice at 0 à Ice at -10:

DS = nC_p*ln(T_f/T_i) = -2.8 J/K

Therefore, DS_tot=DS_sys =5.6 J/K - 44.0 J/K - 2.8 J/K = -41.2 J/K

To find DS_surr, we must find q for each step, add those numbers together and plug into the equation DS_surr = q_rev/T.

Step 1: q_surr=-nC_PDT = -1.5 kJ
Step 2: q_surr=n*DH_fus = 12 kJ
Step 3: q_surr=-nC_pDT = 754 J

q_surr(tot) = 11 kJ, DS_surr=41.8 J/K

DS_univ = DS_sys + DS_surr = 0.6 J/K

7. A 0.590 mol sample of an ideal gas initially at 300 K and 1.50 bar is compressed isothermally to a final pressure of 6.90 bar.  Calculate the change in Gibbs energy for this process.

For an ideal gas, DG = nRT*ln(P₂/P₁) = 2.23 kJ

1.23 Activity

Chem 152 B Activity 1/24/12 – answers will be posted at chem152B.blogspot.com

1. For a chemical reaction involving only the gas phase, entropy is related to the total number of moles on either side of the equation.  A decrease means lower entropy; an increase means higher entropy.

For the reaction

2H₂(g) + O₂(g) à 2H₂O(g)

do you expect the change in entropy to be positive or negative?  Why?  Do you expect the change in enthalpy to be positive or negative?  Why?  (Think about your background knowledge of this reaction.)  What might this tell you about the spontaneity of the reaction?

You would expect the change in entropy to be negative because there are more moles of gas in the reactants than the products.  You would expect the change in enthalpy be negative because we know this reaction to be exothermic – hydrogen is flammable in air.  DS_univ = DS_sys + DS_surr = DS_sys  - DH/T.  DH is negative so we have a negative number minus a negative number.  When temperature is small, DH will outweigh DS and the reaction will be spontaneous (i.e. DS_univ will be positive).  When temperature is large, DS will outweigh DH and the reaction will not be spontaneous (i.e. DS_univ will be negative).

2. For a chemical reaction involving different phases, the production of gas will, in general, increase the entropy much more than the increase in the number of moles of a liquid or solid.

For the reaction

2HNO₃(aq) + Na₂CO₃(s) à 2NaNO₃(aq) + H₂O(l) + CO₂(g)

do you expect the change in entropy to be positive or negative?  Why?

You would expect the change in entropy to be positive since there are more moles of gas in the products than the reactants.

1.17 Activity

1/10 Activity

1/10/12 Activity – Solutions will be posted on blog at chem152B.blogspot.com

1. Argon is compressed adiabatically from an initial volume of 5 L to a final volume of 1 L against a pressure of 10 atm.  Initial temperature is 306K.  Find q, w, delta H, and delta E for this process.  Do the signs of q and w make sense?

To Find w, use w=-PΔV.  ΔV = 1L – 5L = -4L.  P = 10 atm.  w = 40 L*atm.  Be sure to convert to units of Joules!  1 L*atm = 101.3 J.  w = 4052 J   (w is + for compressions.)

q = 0 because this process is adiabatic.  By definition, heat transfer, or q, = 0.  ΔE = q + w.  ΔE = 4052 J

H = E + PV.  Therefore, ΔH = ΔE + Δ(PV).  Δ(PV) must be expanded using the product rule (as in calculus): Δ(PV) = ΔP*V + P*ΔV.  The first term is zero because pressure is not changing.  The second term = -w.  Therefore, ΔH = Δ E –w = 0.  ΔH = 0

2. One mole of argon is compressed from an initial volume of 5 L to a final volume of 1 L against an external pressure and is given sufficient time to thermally equilibrate with its surroundings at 306 K (it is an isothermal process).  Find q, w, delta H, and delta E for this process.  Do the signs of q and w make sense?  Use the equation w_rev=-nRT*ln(V₂/V₁) to solve for w.

By definition, an isothermal process is one where ΔE = 0.  Therefore w = -q.  Using the equation above for w, we get that w = 4095 J.  (w is + for compressions, q is – for heat leaving the system.) ΔH = nC_p ΔT.  We do not need to use C_p here because ΔT = 0 (it is isothermal, meaning temperature does not change).  Therefore, ΔH = 0.

Welcome to Chem 152!

I will post activities here that we do in quiz section.  I will also post any helpful information I find here.  For a review of activities from Chem 142, go to www.chem142.blogspot.com.