Thermodynamics

Calorimetry

100

Calculate the specific heat capacity of given copper of mass 75 g and temperature difference 10^oC if 200 J of heat is lost?

Mass m = 75 g, Temperature difference T = 10^oC, Heat lost Δ Q = 200 J c = qmΔT = 200J/ (75g×10^oC) = 0.267 J/g^oC.

200

How much heat is absorbed when 500. g of water , Cp = 4.184 J/g^oC, goes from 25.0 ^oC to 35.0 ^oC?

Solution: q = cmΔT q= (4.184 J/goC)(500g)(10 ^oC) q= 20920 J

300

A 50.0 g block of glass (Cp = 0.50 J/g^oC) absorbs 333 joules of heat energy. How much does the temperature of the glass rise?

q = cmΔT T= q/cm T= (333J)/ (0.50 J/g^oC x 50g) T= 13.32 ^oC

400

Determine the final temperature when a 25.0g piece of iron at 85.0°C is placed into 75.0grams of water at 20.0°C. The specific heat of iron is 0.450 J/g°C. The specific heat of water is 4.18 J/g°C.

-qmetal=qwater -(mCDT)=mCDT -(mC(Tf-Ti))= mC(Tf-Ti) -(25.0g(0.450J/goC)(Tf-85.0oC))=75.0g(4.18J/goC)(Tf-20.0oC) 956.25-11.25Tf=313.5Tf-6270 7226.25=324.75Tf 7226.25/324.75=Tf 22.3oC=Tf

500

When 20 g of Al (s) at 98 degrees Celsius is placed in 50 g of H2O, the final temperature is 26.5 degrees Celcius. What was theinitial temperature of H2O? Given: CspAl(s) = 0.903 J/g ^oC and cspH20(l) = 4.18 J/g oC

qH2O=−qAl (mH2O)cspH20(l)(ΔT)=(mAl)(CspAl)(ΔT) (50g)(4.18J/goC)(26.5^oC−TiH2O)=−(20g)(0.903J/g^oC)(26.5−98^oC) TiH2O=20^oC

Products/Reactants

100

Does Burning wood illustrates an increase in entropy?

Yes

200

Calculate ΔG for the following reaction at 25^oC. Will the reaction occur spontaneously? NH3(g)+HCl(g)→NH4Cl(s) given for the reaction ΔH=−176.0kJ ΔS=−284.8J/K

calculate ΔG from the formula: ΔG=ΔH−TΔS but first we need to convert the units for ΔS into kJ/K (or convert ΔH into J) and temperature into Kelvin ΔS=−284.8J/K(1kJ1000J)=−0.0284.8kJ/K T=273.15K+25^oC=298K The definition of Gibbs energy can then be used directly ΔG=ΔH−TΔS ΔG=−176.0kJ−(298K)(−0.0284.8kJ/K) ΔG=−176.0kJ−(−84.9kJ) ΔG=−91.1kJ Yes, this reaction is spontaneous at room temperature since ΔG is negative.

300

Calculate ΔH and ΔS for the following reaction and decide in which direction each of these factors will drive the reaction. Compound Hfo(kJ/mol) S�(J/mol-K) N2(g) 0 191.61 H2(g) 0 130.68 NH3(g) -46.11 192.45

The reaction is exothermic ( H� < 0), Ho = Hf^o(products) - Hf^o(reactants) = [2 mol NH3 x 46.11 kJ/mol] - [1 mol N2 x 0 kJ/mol + 3 mol H2 x 0 kJ/mol] = -92.22 kJ The entropy of reaction is unfavorable, S^o = S^o(products) - S^o(reactants) = [2 mol NH3 x 192.45 J/mol-K] - [1 mol N2 x 191.61 J/mol-K + 3 mol H2 x 130.68 J/mol-K] = -198.75 J/K

400

Calculate the standard enthalpy of combustion of the transition of C(s, graphite) → C(s, diamond), given C(s, graphite) + O2 → CO2 ΔHo = -393.5 kJ/mol CO2 → C(s, diamond) + O2 ΔHo = + 395.41 kJ/mol

First we see that both equations are balanced. The enthalpies were given and there is no need to flip an equation around because it is possible to cancel out a couple terms as is. What is left is canceling out the O2 and the CO2 species, writing the overall reaction and then summing the two enthalpies together. C(s, graphite) + O2 → CO2 CO2 → C(s, diamond) + O2 Overall Equation becomes C(s,graphite)→C(s,diamond) Adding the enthalpies gives (-393.5 kJ/mol + 395.41 kJ/mol) = +1.91 kJ/mol. Since the \(\Delta{H}o}\) is positive, the reaction is endothermic.

500

Consider the following reaction: 4 NH3(g) + 5 O2(g) ⇌ 6 H2O(g) + 4 NO(g) ΔG⁰ at 298 K.: H2O= -228.6 kJ/mol, NO = 87.6 kJ/mol NH3= −16.4 kJ/mo , O2 =0.0 kJ/mol Calculate ΔG at 298 K for a mixture of 2.0 bar NH3(g), 1.0 bar O2(g), 1.5 bar H2O(g), and 1.2 bar NO(g).

ΔG⁰ = ∑nΔG⁰f (products) – ∑mΔG⁰f (reactants) ΔG⁰ = [(6 x −228.6 kJ/mol) + (4 x 87.6 kJ/mol)] – [(4 x −16.4 kJ/mol) + (5 x 0.0 kJ/mol)] ΔG⁰ = (-1021.2 kJ/mol) – (-65.6 kJ/mol) ΔG⁰= -955.6 kJ/mol = -9.56 x 105 J/mol 2.ΔG = ΔG⁰ + RT ln Q Q = = 1.5 ΔG = (-9.56 x 105 J/mol) + (8.314 J/mol K)(298 K)ln (1.5) ΔG = (-9.56 x 105 J/mol) + (1.0 x 103 J/mol) ΔG = -9.6 x 105 J/mol

Laws

100

The entropy and enthalpy of a reaction are both negative, is it spontaneous?

Only at low temperatures.

200

A system with __________ enthalpy and _________ entropy will never be spontaneous. (In terms of negative and postive values)

Positive enthalpy and negative entropy

300

Calculate the entropy change at 25°C, in J/K for: 2 SO2(g) + O2(g) → 2 SO3(g) Given the following data: SO2(g) 248.1 J/KC mol O2(g) 205.3 J/KCmol SO3(g) 256.6 J/KC mol

S^o = S^o(products) - S^o(reactants) = [2 mol SO33 x 256.6 J/mol-K] - [2 mol SO2 x 248.1 J/mol-K + 1 mol O2 x 205.3 J/mol-K] = -188.3J/K

400

Calculate ΔH for this reaction: CH4(g) + NH3(g) --> HCN(g) + 3H2(g) given: N2(g) + 3 H2(g) ---> 2 NH3(g) ΔH = -91.8 kJ C(s) + 2 H2(g) ---> CH4(g) ΔH = -74.9 kJ H2(g) + 2 C(s) + N2(g) ---> 2 HCN(g) ΔH = +270.3 kJ

1) Analyze what must happen to each equation: a) first eq ⇒ flip and divide by 2 (puts one NH3 on the reactant side) b) second eq ⇒ flip (puts one CH4 on the reactant side) c) third eq ⇒ divide by 2 (puts one HCN on the product side) 2) rewite all equations with the changes: NH3(g) ---> (1/2)N2(g) + (3/2)H2(g) ΔH = +45.9 kJ <--- note sign change & divide by 2 CH4(g) ---> C(s) + 2 H2(g) ΔH = +74.9 kJ <--- note sign change (1/2)H2(g) + C(s) + (1/2)N2(g) ---> HCN(g) ΔH = +135.15 kJ <--- note divided by 2 3) What cancels when you add the equations: (1/2)N2(g) ⇒ first and third equations C(s) ⇒ second and third equations (1/2)H2(g) on the left side of the third equation cancels out (1/2)H2(g) on the right, leaving a total of 3H2(g) on the right (which is what we want) 4) Calculate the ΔH for our reaction: +45.9 kJ plus +74.9 kJ plus +135.15 = 255.95 kJ = 260. kJ (to three sig figs)

500

Calculate the enthalpy for this reaction: 2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ Given the following thermochemical equations: C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(ℓ) ΔH° = -1299.5 kJ C(s) + O2(g) ---> CO2(g) ΔH° = -393.5 kJ H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -285.8 kJ

1) Determine what we must do to the three given equations to get our target equation: a) first eq: flip it so as to put C2H2 on the product side b) second eq: multiply it by two to get 2C c) third eq: do nothing. We need one H2 on the reactant side and that's what we have. 2) Rewrite all three equations with changes applied: 2CO2(g) + H2O(ℓ) ---> C2H2(g) + (5/2)O2(g) ΔH° = +1299.5 kJ 2C(s) + 2O2(g) ---> 2CO2(g) ΔH° = -787 kJ H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -285.8 kJ Notice that the ΔH values changed as well. 3) Examine what cancels: 2CO2 ⇒ first & second equation H2O ⇒ first & third equation (5/2)O2 ⇒ first & sum of second and third equation 4) Add up ΔH values for our answer: +1299.5 kJ + (-787 kJ) + (-285.8 kJ) = +226.7 kJ

Kinetics

100

A reaction follows the rate law: Rate = k[A]2. Which of the following plots will give a straight line? a. 1/[A] versus 1/time b. [A]2 versus time c. 1/[A] versus time d. ln[A] versus time

c) 1/[A] versus time The “2” exponent means this is a second-order rate law. Second-order rate laws give a straight-line plot for 1/[A] versus t.

200

For the following reaction: NO2 (g) + CO (g) -> NO (g) + CO2 (g), the rate law is: Rate = k[NO2]2. If a small amount of gaseous carbon monoxide (CO) is added to a reaction mixture that was 0.10 molar in NO2 and 0.20 molar in CO, which of the following statements is true? a. Both k and the reaction rate remain the same b. Both k and the reaction rate increase c. Both k and the reaction rate decrease d. Only k increases, the reaction rate will remain the same

a) Both k and the reaction rate remain the same The value k remains the same unless the temperature is changed or a catalyst is added. Only materials that appear in the rate law (NO2) will affect the rate. Adding NO2 would increase the rate while moving it would decrease the rate. CO has no effect on the rate.

300

All factors affect the rates of chemical reaction EXCEPT: a. Temperature b. Catalyst c. Concentration of reactants d. Number of reactants e. Physical state of reactants

d) Number of reactants The temperature, concentration of reactants, physical state of reactants, and presence of a catalyst all affect the rates of chemical reactions.

400

Determine the differential rate law for Zeroth, First, and Second order.

Zeroth Order: Rate = - Δ[A] / Δt = k First Order: Rate = - Δ[A] / Δt = k[A] Second Order: Rate = - Δ[A] / Δt = k[A]²

500

The following mechanism has been proposed for the reaction of CHCl3 with Cl2. Step 1: Cl2 (g) -> 2 Cl (g) fast Step 2: + CHCl3 (g) -> CCl3 (g) + HCl (g) slow Step 3: CCl3 (g) + Cl (g) -> CCl4 (g) fast Which of the following rate laws is consistent with this mechanism? Rate = k[Cl2] Rate = k[CHCl3][Cl2] Rate = k[CHCl3] Rate = k[CHCl3][Cl2]^1/2

d) Rate = k[CHCl3][Cl2]^1/2 The rate law depends on the slow step of the mechanism. The reactants in the slow step are Cl and CHCl3 (one of each). The rate law is first order with respect to each reactant. The Cl is half of the original reactant molecule Cl2, so the [Cl] in the rate law is replaced with [Cl2]^1/2.

Catalyst

100

A substance that speed up the rate of reaction without being consumed in the reaction is __

Catalyst – allows the reaction to proceed more quickly by providing a different mechanism that has a lower activation energy

200

Catalysts that are in the same phase or state of matter as the reactants are ______

Homogenous – they provide an alternate reaction mechanism with a lower activation energy

300

Catalysts that are in a different phase or state of matter from the reactants are _______

Heterogeneous – they lower the activation energy by providing a surface for the reaction and providing a better orientation of one reactant so its reactive site is more easily hit by the other

400

How does a catalyst affect the activation energy of a reacion?

Catalysts lower the activation energy of a reaction.

500

The steps below represent a proposed mechanism for the catalyzed oxidation of CO by O3. Step 1: NO2 (g) + CO (g) -> NO (g) + CO2 (g) Step 2: NO (g) + O3 (g) -> NO2 (g) + O2 (g) What are the overall products of the catalyzed reaction? a. CO2 and O2 b. NO and CO2 c. NO2 and O2 d. NO and O2

a) CO2 and O2 Add the two equations together: NO2 (g) + CO (g) + NO (g) + O3 (g) -> NO (g) + CO2 (g) + NO2 (g) + O2 (g) Then cancel out identical species that appear on opposite sides: CO (g) + O3 (g) -> CO2 (g) + O2 (g)