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Practice Problems in Thermodynamics: Calculating Enthalpy Changes - Prof. Glake A. Hill, Assignments of Chemistry

Jackson State University (JSU)Chemistry
Prof. Glake A. Hill

A collection of practice problems from a thermodynamics textbook, focusing on the calculation of enthalpy changes using various methods such as multiplication by a constant or reversing equations. The problems involve calculating the enthalpy change for different chemical reactions and determining limiting reactants.

Typology: Assignments

Pre 2010

Uploaded on 08/08/2009

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Chapter 6 Practice Problems [Watts, Fall 2008.]
Problem 20
U = q + w
Q = 888 J = 0.888 kJ; w = -7.02 kJ
Note: w is negative since the system does 7.02 kJ of work (on the surroundings). w is the
work done on the system.
U = 0.888 kJ + (-7.02 kJ) = -6.132 kJ
Problem 30
(a) Since the equation has to be multiplied by 8, ΔH is multiplied by 8:H is multiplied by 8:
ΔH is multiplied by 8:H = 8 (-109 kJ) = -872 kJ
(b) The equation is reversed and divided by 2, so the sign of ΔH is multiplied by 8:H changes and ΔH is multiplied by 8:H is
divided by 2:
ΔH is multiplied by 8:H = -(-109 kJ) / 2 = 54.5 kJ
Problem 34
? mol C(s) =
)(1069.2
01.12
10323
4
1
3
sCmol
molg
g
For 323 kg of carbon, the enthalpy change is
kJ
Cmol
kJ
CmolH
64
1017.4
3
8.464
1069.2
Problem 36
CaC2 (s) + 2H2O (l) → C2H2 (g) + Ca(OH)2 (s) ΔH is multiplied by 8:H = -128.0 kJ
We must first find out which of the two reactants is limiting: the question mentions the
amounts of both reactants, and we don’t know at the beginning whether one is in excess
or if the amounts are stoichiometrically equivalent.
pf3
pf4
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Chapter 6 Practice Problems [Watts, Fall 2008.] Problem 20 U = q + w Q = 888 J = 0.888 kJ; w = -7.02 kJ Note: w is negative since the system does 7.02 kJ of work (on the surroundings). w is the work done on the system. U = 0.888 kJ + (-7.02 kJ) = -6.132 kJ Problem 30 (a) Since the equation has to be multiplied by 8, ΔH is multiplied by 8:H is multiplied by 8: ΔH is multiplied by 8:H = 8 (-109 kJ) = -872 kJ (b) The equation is reversed and divided by 2, so the sign of ΔH is multiplied by 8:H changes and ΔH is multiplied by 8:H is divided by 2: ΔH is multiplied by 8:H = -(-109 kJ) / 2 = 54.5 kJ Problem 34 ? mol C(s) = 2.^6910 ( )

  1. 01 (^323 ) 1 3 molC s g mol g     For 323 kg of carbon, the enthalpy change is kJ mol C kJ H^4 mol C 4. 17 106 3
  2. 8   2. 69  10   Problem 36 CaC 2 (s) + 2H 2 O (l) → C 2 H 2 (g) + Ca(OH) 2 (s) ΔH is multiplied by 8:H = -128.0 kJ We must first find out which of the two reactants is limiting: the question mentions the amounts of both reactants, and we don’t know at the beginning whether one is in excess or if the amounts are stoichiometrically equivalent.

Number of moles of CaC 2 = mol g mol g

  1. 6
  2. 10 3500  1  Number of moles of H 2 O = mol g mol g
  3. 4
  4. 02 1250  1  Note: to get a mass of 1250 g for water, the density was taken to be 1 g mL-1, which is a very good approximation. Since 2 moles of H 2 O are needed for each mole of CaC 2 , we see that the water is the limiting reactant. Therefore the ΔH is multiplied by 8:H can be calculated from the number of moles of H 2 O. kJ molH O kJ H molH O 4441. 6 2
  5. 0
  6. 4 2 2     Problem 44 Basic equation: q = msΔH is multiplied by 8:T = CΔH is multiplied by 8:T, where s is the specific heat and C is the heat capacity. ΔH is multiplied by 8:T = 107 oC – 5 oC = 102 oC. C = q / ΔH is multiplied by 8:T = 932 J / 102 oC = 9.14 J oC- Problem 48 The specific heat of sulfur is 0.706 J g-1^ oC-1^ [Table 6.1, page 231]. C g Jg C J ms q T (^) o 94. 0 o
  7. 2 0. 706 2402      1  1  T (^) f  TT i  94. 0  25. 0  119. 0 o C This is the melting point of sulfur: 2402 J is needed to take 36.2 g of sulfur from 25.0 oC to its melting point. Problem 50 qmetal + qwater = 0 qwater = msΔH is multiplied by 8:T = (28.0 g) (4.180 J g-1^ oC-1) (21.23 oC – 19.73 oC) = 176 J qmetal = -176 J = msΔH is multiplied by 8:T

To get the heat per gram, divide by 0.828 g: qrxn (per gram) = -35.8 kJ / 0.828 g = -43. kJ/g. Problem 66 N 2 (g) + 2O 2 (g) → N 2 O 4 (g) ΔH is multiplied by 8:H = 9.2 kJ (1) N 2 (g) + 2O 2 (g) → 2NO 2 (g) ΔH is multiplied by 8:H = 33.2 kJ (2) Since we want NO 2 to be a reactant in the reaction, reverse the second equation 2NO 2 (g) → N 2 (g) + 2O 2 (g) ΔH is multiplied by 8:H = -33.2 kJ (3) Add reactions (1) and (3). This leads to 2NO 2 (g) → N 2 O 4 (g) [The N 2 and O 2 cancel out] By Hess’s law, ΔH is multiplied by 8:H for this reaction is equal to 9.2 + (-33.2) = -24.0 kJ. Problem 68 C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O (l) ΔH is multiplied by 8:H = -2219.9 kJ (1) CO (g) + ½O 2 (g) → CO 2 (g) ΔH is multiplied by 8:H = -283.0 kJ (2) Since we want 3CO(g) in the product, multiply (2) by 3 and reverse it: 3CO 2 (g) → 3CO (g) + (3/2) O 2 (g) ΔH is multiplied by 8:H = 849.0 kJ (3) Add (1) and (3) together: C 3 H 8 (g) + 5O 2 (g) + 3CO 2 (g) → 3CO 2 (g) + 4H 2 O (l) + 3CO (g) + (3/2) O 2 (g) The CO 2 cancels. There is also a partial cancellation of the O 2. The above reaction is equivalent to C 3 H 8 (g) + (7/2) O 2 (g) → 4H 2 O (l) + 3CO (g) The ΔH is multiplied by 8:H is the sum of the ΔH is multiplied by 8:H’s of (1) and (3): -2219.9 + 849.0 = -1370.9 kJ.

Problem 72 Obtain the standard heats of formation from the back of the textbook. Pay careful attention to physical states. NH 3 (g) + HCl (g) → NH 4 Cl (s) ΔH is multiplied by 8:Ho^ = -314.4 – [(-46.11) + (-92.31)] = -175.98 kJ NH 3 (g) + HNO 3 ( l ) → NH 4 NO 3 (s) ΔH is multiplied by 8:Ho^ = -365.6 – [(-46.11) + (-173.2)] = -146.29 kJ Zn (s) + 2 HCl (g) → ZnCl 2 (s) + H 2 (g) ΔH is multiplied by 8:Ho^ = -415.1 – 2(-92.31) = -230.48 kJ FeO (s) + CO (g) → Fe (s) + CO 2 (g) ΔH is multiplied by 8:Ho^ = -393.5 – [(-110.5) + (-272)] = -11 kJ Problem 74 We have to solve for the standard enthalpy of formation for sucrose. We use the given ΔH is multiplied by 8:Ho^ and look up the ΔH is multiplied by 8:Hof of the other chemicals. ΔH is multiplied by 8:Ho^ = 12ΔH is multiplied by 8:Hof (CO 2 (g)) + 11ΔH is multiplied by 8:Hof (H 2 O ( l )) – [ΔH is multiplied by 8:Hof (C 12 H 22 O 11 (s)) + 12ΔH is multiplied by 8:Hof (O 2 (g))] ΔH is multiplied by 8:Hof (C 12 H 22 O 11 (s)) = 12ΔH is multiplied by 8:Hof (CO 2 (g)) + 11ΔH is multiplied by 8:Hof (H 2 O ( l )) – [12ΔH is multiplied by 8:Hof (O 2 (g)) + ΔH is multiplied by 8:Ho] = 12(-393.5) + 11(-285.8) – (-5650) = -4722 -3143.8 + 5650 = -2215.8 kJ mol- [Note: strictly there is a division by 1 mol, hence the final units of kJ mol-1. Think about it.]