Thermodynamics Problems

THERMODYNAMICS PROBLEMS 1. Calculate H rxn for the following reaction at 25 °C. 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) kJ kJ H f [NO(g)] = 90.4 H f [H2O(g)] = -241.8 mol mol H f [NH3(g)] = -46.3 kJ mol kJ kJ kJ kJ ) + (6)(-241.8 ) - [(4)(-46.3 ) + (5)(0)] = -904.0 mol mol mol mol H rxn = (4)(90.4 If you include moles as units for your coefficients: H rxn = -904.0 kJ 2. Calculate the entropy change for 2Na(s) + Cl2(g)  2NaCl(s) S values: Na(s) = 51.05 J/mol•K, Cl2(g) = 223.0 J/mol•K, NaCl(s) = 72.38 J/mol•K  S rxn = (2)  72.38  J J  J  J     - (2)  51.05  - (1)  223.0  = -180.3 K  mol  K  mol K  mol     K  mol If you include moles as units for your coefficients: S rxn = -180.3 J K 3. a) Calculate Ggiven H = -227 kJ, S = -309 J/K, T = 1450 K.   309 J  1kJ   221 kJ   K  1000 J  G = H - TS G   227 kJ  (1450 K ) b) Is this process spontaneous at this temperature? If not, calculate the temperature (in oC) at which this reaction becomes spontaneous. No, it is not spontaneous at 1450 K (G is +) Set G = 0: T= 0 = H - TS  T= 227 kJ = 735 K -273 = 462 oC   J  1kJ     309  K  1000 J   H S  T at equilibrium H and S are both negative, so it is spontaneous at temperatures below 462 oC. Spontaneous at T < 462 oC 3. Gfo in kJ/mol: CaO(s) -604.2 Ca(OH)2(s) -896.8 H2O(g) -228.6 H2O(l) -237.2 Calculate G for these reactions and predict whether they will be spontaneous or not. o a) 2H2O(g)  2H2(g) + O2(g)  G = -(2)   228.6  kJ kJ  ( or 457.2 kJ) +G so it is not spontaneous  = 457.2 mol  mol b) CaO(s) + H2O(l)  Ca(OH)2(s)  G = (1)   896.8  kJ kJ  kJ  kJ    (or -55.4 kJ )  - (1)   604.2  - (1)   237.2  = -55.4 mol  mol mol     mol b is spontaneous under standard state conditions (G is -) Page 1 of 2 4. Go is -24.7 kJ/mol for the formation of methanol. C(s) + ½ O2(g) + 2H2(g)  CH3OH(g) Calculate the equilibrium constant, K, at 25  C for this reaction. K=e  G RT =e kJ    24.7  mol    J  1kJ    298 K   8.314  K mol   1000 J   = e9.969 = 2.14104 5. a) At 25 C, Ka for acetic acid is 1.810-5. Predict the sign of Go at 25 oC for CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO-(aq). Calculate Go at 25 C. K < 1 so G will be + G = -RT ln K = - (8.314 J )(298 K)(ln 1.8 x 10-5) = 2.7104 K  mol J kJ or 27 mol mol b) Calculate G at 25 C for the acetic acid equilibrium reaction, when [H3O+] = 0.020 M, [CH3COO-] = 0.010 M, and [CH3COOH] = 0.10 M. (Use Go from part b.) G = G + RTlnQ Q= [H 3O  ][CH3COO ] [CH3COOH] G = 27 kJ  [0.020][0.010]  kJ  + (8.31410-3 )(298 K)  ln [0.10] K  mol mol   G = 27 kJ kJ kJ kJ + (2.478 )(ln 2.010-3) = 27 + (2.478 )(-6.215) mol mol mol mol G = 27 kJ kJ - 15.4 mol mol G = 12 kJ or 1.2104 mol J mol  Since Q = 2.010-3 > K = 1.8x10-5, the reaction shifts left This is consistent with the +G value which tells us the reaction is not spontaneous in the forward direction. 6. ∆G˚ for the reaction H2(g) + I2(g)  2 HI(g) is 2.60 kJ/mol at 25°C. In one experiment, the initial pressures are PH2 = 4.3 atm, PI2 = 0.34 atm, and PHI = 0.23 atm. Calculate G and predict the direction that this reaction will proceed. G = G + RTlnQ Q = PHI2 PH PI 2 G = 2.60 G = 2.60 2 kJ  kJ (0.23) 2  + (8.31410-3 )(298 K)  ln  K  mol mol  ( 4.3)(0.34)  kJ kJ kJ kJ kJ + (2.478 )(ln 0.0362) = 2.60 + -8.22 = -5.62 mol mol mol mol mol Since G is negative it will be spontaneous in the forward direction Page 2 of 2