Name __Mr. Perfect________________________________________ Date _____Sp 17____________
1. In a first-order decomposition reaction, 50% of a compound (A) decomposes in 10.5 min.
A→B
(a) Calculate the rate constant for this reaction. (5 pts)
𝑘=
ln 2
0.693
=
= 𝟎. 𝟎𝟔𝟔 𝒎𝒊𝒏−𝟏
𝑡1/2 10.5 𝑚𝑖𝑛
(b) How long will it take for 75% of the compound to decompose? (5 pts)
75 % decomposed thus 25 % remains
𝑙𝑛
0.25[𝐴]0
= −𝑘𝑡
[𝐴]0
𝑜𝑟
𝑡=
−1.39
= 𝟐𝟏 𝒎𝒊𝒏
−0.066 𝑚𝑖𝑛−1
2. For the following reaction:
4A + 3B → 2C
Trial
1
2
3
4
[A] M
0.10
0.30
0.10
0.30
[B] M
0.10
0.10
0.20
0.20
Initial Rate (M/s)
5.0
45.0
10.0
90.0
Calculate the orders for A and B and write the rate law for this reaction. Also, calculate the rate
constant for this reaction. Show work to receive full credit. (10 pts)
𝑟𝑎𝑡𝑒3 10.0 𝑘(0.1)𝑥 (0.2)𝑦
=
=
𝑟𝑎𝑡𝑒1
5.0
𝑘(0.1)𝑥 (0.1)𝑦
𝑟𝑎𝑡𝑒4 90.0 𝑘(0.3)𝑥 (0.2)𝑦
=
=
𝑟𝑎𝑡𝑒3 10.0 𝑘(0.1)𝑥 (0.2)𝑦
Rate Law:
2 = 2𝑦
9 = 3𝑥
𝑦 = 1 𝑓𝑖𝑟𝑠𝑡 − 𝑜𝑟𝑑𝑒𝑟 𝑖𝑛 𝐵
𝑥 = 2 𝑠𝑒𝑐𝑜𝑛𝑑 − 𝑜𝑟𝑑𝑒𝑟 𝑖𝑛 𝐴
rate = k[A]2[B]
𝑘=
𝑟𝑎𝑡𝑒
5.0 𝑀/𝑠
=
= 𝟓𝟎𝟎𝟎 𝑴−𝟐 𝒔−𝟏
2
[𝐴] [𝐵] [0.1]2 [0.1]
Chemistry 102 Exam 1
Name __Mr. Perfect________________________________________ Date _____Sp 17____________
3. Calculate the grams of water that can form when 5.2 g of methane, CH4, is reacted with an
excess of O2. The molar mass of methane is 16 g/mol and the molar mass of water is 18 g/mol.
Use the following rate expressions to write a balanced equation for the reaction. (10 pts)
Rate Expressions:
𝑟𝑎𝑡𝑒 = −
∆[𝐶𝐻4 ]
1 ∆[𝑂2 ] 1 ∆[𝐻2 𝑂] ∆[𝐶𝑂2 ]
=−
=
=
∆𝑡
2 ∆𝑡
2 ∆𝑡
∆𝑡
Balanced Equation:
CH4 + 2O2 → 2H2O + CO2
5.2 𝑔 𝐶𝐻4 ×
𝑚𝑜𝑙 2 𝑚𝑜𝑙 𝐻2 𝑂 18 𝑔
×
×
= 𝟏𝟏. 𝟕 𝒈 𝑯𝟐 𝑶
16 𝑔 1 𝑚𝑜𝑙 𝐶𝐻4 𝑚𝑜𝑙
4. The rate constant of a reaction is 4.7 x 10-3 s-1 at 25 °C, and the activation energy is 33.6
kJ/mol. What is the rate constant k at 75 °C? (10 pts)
2 Point Form of the Arrhenius Equation
𝑙𝑛
𝑘2 𝐸𝑎 𝑇2 − 𝑇1
=
(
)
𝑘1
𝑅 𝑇1 𝑇2
(R = 8.314 J/molK)
Convert temperature to Kelvin:
T1 = 25 + 273 = 298 K
𝑙𝑛
T2 = 75 + 273 = 348 K
𝑘2
33.6 𝑥 103 𝐽/𝑚𝑜𝑙
348 𝐾 − 298𝐾)
=
=(
)
−3
−1
(4.7 𝑥 10 𝑠 ) 8.314 𝐽/𝑚𝑜𝑙 ∙ 𝐾
298𝐾 ∙ 348𝐾
𝑙𝑛
𝑘2
= 1.95
(4.7 𝑥 10−3 𝑠 −1 )
lnk2 – (-5.36) = 1.95
or
k2 = e-3.41 = 3.3 x 10-2s-1
Chemistry 102 Exam 1
lnk2 = -3.41
Name __Mr. Perfect________________________________________ Date _____Sp 17____________
5. At 100 °C, Kp = 60.6 for the following reaction:
2NOBr(g) ⇄ 2NO(g) + Br2(g)
In a given experiment, 0.10 atm of each component is placed in a container. Is the system at
equilibrium? If not, in which direction will the reaction proceed? (10 pts)
(0. 1)2 (0.1)
𝑄𝑝 =
= 0.1
(0.1)2
Qp < Kp
Reaction will proceed forward (left to right).
6. In an analysis of interhalogen reactivity, 0.1 M of ICl was placed into a flask where it
decomposed at high temperature in the following reaction:
2ICl(g) ⇄ I2(g) + Cl2(g)
Calculate the equilibrium concentrations of I2, Cl2, and ICl if Kc = 0.11 at this T. (15 pts)
2ICl(g) ⇄ I2(g) + Cl2(g)
I
0.1
0
0
C
-2x
+x
+x
E
0.1-2x
x
x
2
2
𝑥
𝑥
𝐾=
≈
= 0.110
𝑥 = 0.105 𝑀 𝑁𝑜𝑡 𝑉𝑎𝑙𝑖𝑑 (𝑙𝑎𝑟𝑔𝑒𝑟 𝑡ℎ𝑎𝑛 𝑖𝑛𝑖𝑡𝑖𝑎𝑙)
0.1 − 2𝑥 0.1
Solve quadratic:
x2 + 0.22x – 0.011 = 0
−(0.22) ± √(0.22)2 − 4(1)(−0.011)
𝑥 = −0.262 𝑜𝑟 𝑥 = 0.042
2(1)
[ICl] = 0.1 - 2(0.042) = 0.016 M
[I2] = 0.042 M and [Cl2] = 0.042 M
7. Consider the following overall reaction:
A+ B→ G
A plot of ln [A] vs t gives a straight line. Determine the slow step in the following proposed
mechanism. Clearly state the order of each reactant and write the overall rate law. (10 pts)
Linear equation for a first-order reaction: ln[A]t = -kt + ln[A]0
Proposed Mechanism
Step 1: A → Z
(slow step) rate = k[A]
Step 2: Z + B ⇄ ZB
Step 3: ZB → G
Overall reaction: A + B → G
overall rate law:
rate = k[A][B]0 = k[A]
𝑥=
[A] first-order
[B] zero-order
Chemistry 102 Exam 1
Name __Mr. Perfect________________________________________ Date _____Sp 17____________
8. Hydrogen sulfide decomposes according to the following reaction with a Kc = 9.30 x 10-8 at
700 °C.
3rd Half-life: 12.5 % of the initial remains thus 87.5 % was lost.
2H2S(g) ⇄ 2H2(g) + S2(g)
If 0.45 mol of H2S is placed into a 3.0 liter container, calculate the equilibrium concentrations of
the reactants and products at the 3rd half-life. (10 pts)
2H2S(g) ⇄ 2H2(g) + S2(g)
I 0.15
0
0
C
-x
+2x
+x
E 0.15-x
2x
x
rd
x = 0.15(0.875) =0.13 M lost in the reaction at the 3 half-life
[H2S] = 0.15 – 0.13 =0.02 M ;
[H2] = 2(0.13) =0.26 M ;
[S2] = 0.13 M
9. A Certain reaction has the general form:
A → B
The plot of 1n[A] vs time resulted in a straight line with a slope equal to 3.6 x 10-2 s-1. The initial
concentration of A is 2.8 x 10-3 M.
[A]t = -kt + [A]o
Some useful equations:
ln[A]t = -kt + ln[A]o
1/[A]t = kt + 1/[A]o
A) Determine the rate law for this reaction. (5 pts)
First-order rate law:
rate = k[A]
B) Calculate the half-life for this reaction. (5 pts)
𝑡1/2 =
𝑙𝑛2
0.693
=
= 𝟏𝟗. 𝟑 𝒔
𝑘
3.6 𝑥 10−2 𝑠 −1
C) How much time (seconds) is required for the initial concentration of A to decrease to 7.0
x 10-4 M? (5 pts)
ln (7 x 10-4) = -(3.6 x 10-2s-1)(t) + ln (2.8 x 10-3)
t = 38.3 s
10. Extra Credit. Which direction will the following reaction shift if the pressure of the reaction
mixture is doubled? (5 pts)
2O3(g) ⇄ 3O2(g)
The reaction will shift to the left (less moles of gas).
Chemistry 102 Exam 1