AP CHEMISTRY
Week 5 Problem Set
Solutions
, AP Chemistry Week 5
Problem Set Solution
Multiple Choice Questions
Question 1: Identify the catalyst in this reaction mechanism.
𝑂3 + 𝐶𝑙 → 𝑂2 + 𝐶𝑙𝑂
𝐶𝑙𝑂 + 𝑂 → 𝐶𝑙 + 𝑂2
A) 𝐶𝑙
B) 𝐶𝑙𝑂
C) 𝑂
D) 𝑂2
Answer: A. A catalyse is used in one step and re-produced in another step.
Question 2: Methane combusts with oxygen to yield carbon dioxide and water vapor:
CH4 + 2O2 → CO2 + 2H2O
If methane is consumed at 2.79mol s–1, what is the rate of change in the
concentrations of carbon dioxide and oxygen?
A) +2.79mol s–1 CO2 and +5.58mol s–1 O2
B) –2.79mol s–1 CO2 and +5.58mol s–1 O2
C) +5.58mol s–1 CO2 and –5.58mol s–1 O2
D) +2.79mol s–1 CO2 and –2.79mol s–1 O2
E) +2.79mol s–1 CO2 and –5.58mol s–1 O2
Answer: E. The reaction equation makes clear that for each mole of methane consumed,
2 moles of oxygen gas are consumed and 1 mole of carbon dioxide is produced. So
d[CH4]/dt = 0.5d[O2]/dt = –d[CO2]/dt
This means that the disappearance of 2.79mol s–1 of methane corresponds to the
appearance (positive change) of 2.79mol s–1 of CO2 and the disappearance (negative
change) of 2 (– 2.79) = 5.58mol s–1 of O2. The concentration of methane changes at half
the rate of oxygen, so the concentration oxygen changes at twice the rate of methane.
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Week 5 Problem Set
Solutions
, AP Chemistry Week 5
Problem Set Solution
Multiple Choice Questions
Question 1: Identify the catalyst in this reaction mechanism.
𝑂3 + 𝐶𝑙 → 𝑂2 + 𝐶𝑙𝑂
𝐶𝑙𝑂 + 𝑂 → 𝐶𝑙 + 𝑂2
A) 𝐶𝑙
B) 𝐶𝑙𝑂
C) 𝑂
D) 𝑂2
Answer: A. A catalyse is used in one step and re-produced in another step.
Question 2: Methane combusts with oxygen to yield carbon dioxide and water vapor:
CH4 + 2O2 → CO2 + 2H2O
If methane is consumed at 2.79mol s–1, what is the rate of change in the
concentrations of carbon dioxide and oxygen?
A) +2.79mol s–1 CO2 and +5.58mol s–1 O2
B) –2.79mol s–1 CO2 and +5.58mol s–1 O2
C) +5.58mol s–1 CO2 and –5.58mol s–1 O2
D) +2.79mol s–1 CO2 and –2.79mol s–1 O2
E) +2.79mol s–1 CO2 and –5.58mol s–1 O2
Answer: E. The reaction equation makes clear that for each mole of methane consumed,
2 moles of oxygen gas are consumed and 1 mole of carbon dioxide is produced. So
d[CH4]/dt = 0.5d[O2]/dt = –d[CO2]/dt
This means that the disappearance of 2.79mol s–1 of methane corresponds to the
appearance (positive change) of 2.79mol s–1 of CO2 and the disappearance (negative
change) of 2 (– 2.79) = 5.58mol s–1 of O2. The concentration of methane changes at half
the rate of oxygen, so the concentration oxygen changes at twice the rate of methane.
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