1. Calculate the enthalpy change for the reaction as written in the balanced chemical equation:

DH¤ = Sm(i)Df(i)

DH¤(C8H18)

= (-2)(-255.1) + (-25)(0) + (+16)(-393.5) + (+18)(-285.8)

= -10,916 kJ

DH¤(C2H6O)

= (-1)(-277.0) + (-3)(0) + (+2)(-393.5) + (+3)(-285.8)

= -1367.4 kJ

2. Calculate the enthalpy change per litre:

C8H18:

{-10,986 kJ / 2 mol}*{ 1 mol / 114.22 g} * {719.0 g / L}

= -34,360 kJ/L

C2H6

{-1367.4 kJ / mol}*{ 1 mol / 46.07 g} * {789.4 g / L}

=-23,430 kJ/L

Thus gasoline releases more energy per litre than ethanol. Adding ethanol decreases the amount of energy available per L.

2a.

At 298 K: ln{K} = -DG¤ / {R*298} = -(-1.756) / {8.314*298}

= 0.7088

So K = 2.03.

Note that a different K is obtained if DG = DH - TDS is used.

 Br2(l) + Cl2(g) ƒ 2 BrCl(g) I excess 100 kPa 0 C -x +2x E excess 100 - x 2x

K = a(BrCl)2 / {a(Br2) * a (Cl2)}

= {P(BrCl) / P¤}2 / {1 * (P(Cl2) / P¤)}

= {2x/100}2 / {(100 - x)/100}

=2.03

Solving: x = 50.3

Thus P(Cl2) = 49.7 kPa, P(BrCl) = 100.6 kPa.

2b. Consider the same equilibrium

Br2(l) + Cl2(g) ƒ 2 BrCl(g).

and using the le Chãtelier Principle, fill in the following table with L(eft), R(ight), N(o change) or U(p), D(own), N(o change).

 Change imposed on equilibrium mixture Effect on equilibrium position (L, R, N) Effect on value of K (U, D, N) Add Br2(l) Add BrCl(g) Decrease volume Add a catalyst Add inert He(g) at constant volume Heat the mixture +10 K

4. The following parts of this question refer to the equilibrium at 298 K:

2 NO(g) + 2 H2(g) ƒ N2(g) + 2 H2O(g) DH¤ = -664.4 kJ.

Other datum: Df(H2O(g)) = -241.8 kJ.

 a. Calculate DE¤ for the reaction.

DH¤ = DE¤ + PDV = DE¤ + RT*Dn

-664.4 = DE¤ + *.314 * 298 * (-1)

DE¤ = -661.9 kJ

 b. Calculate Df(NO(g)).

DH¤ = -664.4 kJ = Sm(i)Df(i)

= (-2)(Df(NO)) + (-2)(0) + (+1)(0) + (+2)(-241.8)

Df(NO) = +90.4 kJ

 c. Predict (do not calculate) the sign (+, -, 0) for DS¤, giving your reasons.

Since Dn for the gases is -1, the system is becoming more ordered, so DS¤ is predicted to be negative.

 d. Upon heating predict the direction of movement of the equilibrium.

Reaction will move in the endothermic direction when heated, in this case to the left.

 e. A kinetic investigation of the reaction showed that the rate halved as [H2] halved, and quadrupled when [NO] doubled. Write down the experimental rate law this data gives.

rate = k [H2] [ NO]2

 f. At a certain temperature, T, Kc = 3000 L.mol-1. Calculate Kp, and K (activities K) at T.

Kp = P P(i)m (i) = P { [i]m(i) * (RT)m(i)}

= P { [i]m(i) } * (RT)Sm(i)

= Kc * (RT)Dn

So Kp = 3000 * (RT)-1

K = P a(i)m(i) = P { (P(i) / P¤)m(i)

= P { P(i)}m(i) * {1/P¤}Sm(i)

= Kp * (P¤)-Dn

So K = 3000 * (RT)-1 * 100+1 = 300000 * (RT)-1

5

ln K = {DS¤ / R} - {DH¤ / RT}

ln {K2 / K1} = {DH¤ / R} { 1/T1 - 1/T2}

Different pairs give slightly different answers for DH¤.

I used items 1 and 2 in the list:

ln { 11.0 / 16.9} = {DH¤ / 8.314} { 1/270.1 - 1/284.8}

DH¤ = -18,680 J

Substitute this in top equation gives DS¤ = -45.65 J/K

Substitute these two values into: D298 = DH¤ - 298*D

gives D298 = -5076 J