Ch. 10

Grismer #1,3,5,7

Based on the global water balance, 97% of the earth’s water is found in the oceans and is salty.

Of the remaining 3%, 77% of the fresh water is found in the frozen ice caps and is unavailable to man.

3.

First-order reaction

d[A] / dt = -k [A]

where [Ao] is the original concentration and at 40% completion [A] = 0.6[Ao]

Separate variables and integrate

∫ d[A] / [A] = -k ∫ dt

(ln[A] – ln [Ao] ) = ln [A] / [Ao] = -kt

Solve for k

k = -(1 / t) ln (0.6) = -0.02 ln (0.6) = 0.01 /min

At 80% completion

t = -(1 / k) ln (0.2) = -100 ln (0.2) = 161 min

5.

Using table 2.4 (p. 39) and D G° = -RT ln k we have

D G° = [-19 – 37.6] – [-56.69 – 6.37] = 6.46 kcal/mole

(order as ammonium, hydroxide, water and ammonia)

(a)

K = exp (-6.46 / RT)

K = exp [- (6.46 kcal * 1000 cal/kcal) / (1.987 cal/mole K )(298 K)] = 1.83 * 10 ^–5

(b)

For reaction direction we use Q and pH = 8 implies pOH = 6 or [OH] = 10 ^–6

Q = [NH₄][OH] / [H₂0][NH₃] = [10 ^–4][10 ^–6] / [1] [2 * 10 ^–8] = 0.005

Q/K = 0.005 / 1.83 * 10 ^–5 = 273 and D G >0

Therefore the reaction is going in the opposite direction

NH_4(aq) + OH ® NH_3(aq) + H₂O

7.

Again, use the Q/K relationship

Q = [HCO₃][H] / [CO₂] = [10 ^–5] [10 ^–5] / [10 ^–5] = [10 ^–5]

From Table 10.11, K = 1.51 * 10 ^–8

Q / K = 10 ^–5 / 1.51 * 10 ^–8 = 662 and D G >0

Therefore, the reaction is going in the opposite direction, that is,

H ⁺ + HCO₃ ® H₂O + CO_{2 (aq)}