ANSWERS TO EXERCISES AND REVISION QUESTIONS 443Unit 14
Exercises
14A
(i)NO disappears at twice the rate that H 2 disappears
0.02 mol dm–^3 s–^1.
(ii)Nitrous oxide appears at the same rate at which H 2
disappears0.01 mol dm–^3 s–^1.
(iii)Water appears at the same rate at which H 2
disappears0.01 mol dm–^3 s–^1.
14Binitial rate 0.60 10 0.100.050 mol dm–^3 s–^1rate at 25 s 0.27 30 0.154.0 10 –^3 mol dm–^3 s–^1(Your answers might be slightly different due to the
difficulty of drawing tangents.)
14Cinitial rate 12 000 36 400 ^00
0.33 cm^3 of O 2 per secondrate at 50 000 s 10 132 ^5700 4.432^103
60 000 0 60 000
0.074 cm^3 of O 2 per second
14D
The cut divides the marble into two rectangular blocks of
dimensions 0.5 1 1 cm. The surface area of each
block is (0.5 1 1 0.50.50.5)4 cm^3. The
total surface area is the sum of the surface area for both
blocks8 cm^3. This is 33% greater than the surface
area of the undividedblock (6 cm^3 ). Dividing the block
into even smaller pieces greatly increases the total
surface area available for attack by the acid.
14E
(i) 100 20 °C 80 °C so that the increase in rate is
2 2 2 2 2 2 2 2 (i.e. 2^8 )256 times.
(ii)The heat generated by the exothermic reaction raises
the temperature of the reaction mixture, so increasing its
rate.
14F
(i)Nails have a lower surface area than powder.
(ii)(a) Water dilutes the concentration of acid, so
reducing the number of acid–zinc collisions per second.
(b) Ice cools the mixture, thereby reducing reaction rate.
(c) Sodium carbonate reacts with the acid, so drastic
ally reducing the concentration of hydrogen ions in
solution.
14G
(i) 1 (ii) 2 (iii) 1 (iv)3.
14H
(i)Third order.(ii)Rate 7000 (0.060)^3 1.5
mol dm–^3 s–^1.14I
(i)Second order.
(ii)
k initial rate 4.0 ^10 –^6
[C 12 H 22 O 11 ][H+](0.20)(0.10)
2.0 10 –^4 mol–^1 dm^3 s–^1
14J
Rate of reaction k[M][Y]^0 k[M]
14K
rate of reactionk[Br–(aq)]x[BrO 3 – (aq)]y[H+(aq)]z
Comparison of experiments 1 and 2 shows that doubling
[Br–(aq)] doubles the initial rate; therefore,x1.
Comparing experiments 1 and 3 shows that increasing
[BrO 3 – (aq)] by a factor of four increases the initial rate by
four; therefore, y1. Comparing experiments 1 and 4
shows that doubling H+(aq) increases the rate by four;
therefore,z2 and the rate expression is
ratek[Br–(aq)] [BrO 3 – (aq)] [H+(aq)]^2
14L
10 h 10 60 60 36 000 s.
[A]t[A] 0 e–kt0.01 e–(6.5^10 –^6 36 000)
0.01 0.7910.0079 mol dm–^3.
14Mt (^12) ⁄0.693k 9.60.693 10 – 3 72 s
14N
t 1 0.693
⁄ 2 5.73 10 – 7 1 209 424 s (14.0 days).
42 days is three half-lives. During this period the
concentration of pesticide will have fallen to one-eighth
of its initial concentration, i.e. from 0.10 mg dm–^3 to
0.0125 mg dm–^3.
Revision questions
14.1See text.
14.2(i).
14.3 (i)II (ii)I.
14.4 (i) 1
(ii)initial rate
2.00 10 –^5 0.1002.00 10 –^6 mol dm–^3 s–^1
(iii)rate2.00 10 –^5 0.05231.05 10 –^6 mol
dm–^3 s–^1
(iv)half life 0.693/2.00 10 –5
3465 s 9.6 h
14.5 (i) 1
(ii)plot as follows:
(iii)initial rate 0.100.0
3.3 60 60
8.4 10 –^6 mol dm–^3 s–^1
(iv)initial rate k[N 2 O 5 (g)] 0
k(65 °C)8.4 10 –^6 /0.108.4 10 –^5 s–^1
(v)
t 0.693
1 ⁄ 2 8.4 10 – 5 8300 s (2.3 h)
(Inspection of the graph and table confirms that the
reaction half-life is roughly 2 h.)
14.6 (i)Consumption of hydroxide ion is negligible and
rate of reaction depends upon the concentration of the
bromomethane only:
rate of reaction k[CH 3 Br(l)]
withkk[OH–(aq)]. If the bromomethane were in
excess, the reaction would also be pseudo first order and
the rate expression would be
rate of reaction k [OH–(aq)]
withk k[CH 3 Br(l)]
(ii) (a)Initial rate of reactionk[CH 3 Br(l)][OH–(aq)]
3.0 10 –^4 1.0
10 –^2 1.0 10 –^4
3.0 10 –^10 mol dm–^3 s–^1
(b)pseudo-rate constant kk[OH–(aq)]
3.0 10 –^4 1.0 10 –^2 3.0 10 –^6 s–^1
(c)pseudo half-life of reaction 0.693 / 3.0 10 –^6
2.3 105 s.
14.7 (i)
ratek[ClO]x[NO 2 ]y[N 2 ]z
Comparison of experiments 1 and 2 shows that halving
the concentration of ClO halves the initial rate; therefore,
x1. Experiments 1 and 3 show that doubling [NO 2 ]
doubles the rate; therefore, y1. Similarly, comparison
of experiments 2 and 4 shows that doubling [N 2 ] doubles
the initial rate; therefore, z1. The rate expression
becomes
ratek[ClO][NO 2 ][N 2 ]
(ii)From the data of experiment 1,
k initial rate
[ClO][NO 2 ][N 2 ]
k 3.5^10 –^4
[1.0 10 –^5 ][2.0 10 –^5 ][3.0 10 –^5 ]
k5.8 1010 mol–^2 dm^6 s–^1
From the data of the remaining experiments:
experiment 2: k6.0 1010
experiment 3: k5.9 1010
experiment 4: k6.0 1010
Averagek(298 K) 5.9 1010 mol–^2 dm^6 s–^1.
(iii)If the reaction between Cl0, NO 2 and N 2 were
elementary, we would expect the rate expression to be
K[Cl0][NO 2 ][N 2 ], as above. The rate expression is
thereforeconsistentwith a one-step mechanism.
14.8
14.9First derive the expression
ln([A][A]^0 )kt
t
as in the text. Symbolize the time at which 99% of the
concentration of A has disappeared as t0.99. After this
time, the concentration of A will have fallen to one
hundredth of its initial value:
[A] [A]^0
0.99 100
Substitution gives
ln [A]^0
([A] 0 /100)ln (100) kt0.99
Since ln (100) 4.605,
4.605kt0.99
ort0.994.605k
0230_000118_2 9 _Ans.qxd 3/2/06 2:22 pm Page 443