26.0k views
5 votes
Consider the combustion of ethylene: C₂H₄(g) + 3 O₂(g) → 2 CO₂(g) + 2 H₂O(g) if the concentration of C₂H₄ is decreasing at the rate of 3.6×10⁻² m/s, what is the rate of change in the concentration of CO₂?

Express the rate in molarity per second to two significant digits.

User Amgad
by
7.3k points

1 Answer

4 votes

Final answer:

The rate of change in concentration of CO₂ when C₂H₄ is consumed at 3.6×10⁻² M/s is 7.2×10⁻² M/s. This is determined by doubling the consumption rate of C₂H₄ based on the stoichiometry of the reaction.

Step-by-step explanation:

The student has asked: If the concentration of C₂H₄ is decreasing at the rate of 3.6×10⁻² M/s, what is the rate of change in the concentration of CO₂? To solve this, we analyze the stoichiometry of the balanced combustion reaction of ethylene which is:

C₂H₄(g) + 3 O₂(g) → 2 CO₂(g) + 2 H₂O(g).

From the coefficients, we see that for every 1 molecule of C₂H₄ consumed, 2 molecules of CO₂ are produced. This means that the rate of production of CO₂ is twice the rate of consumption of C₂H₄.

Therefore, if C₂H₄ is consumed at 3.6×10⁻² M/s, the CO₂ is produced at a rate of 7.2×10⁻² M/s, since 2 times 3.6×10⁻² equals 7.2×10⁻². We express the rate in molarity per second at two significant digits as 7.2×10⁻² M/s.

User Beer
by
8.3k points