In a chemistry class, a student follows the breakdown of a gas in a closed vessel…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
In a chemistry class, a student follows the breakdown of a gas in a closed vessel. The concentration of the gas, in mol/L, depends on the time in seconds. A data book says that the breakdown is a second-order reaction: the rate of fall of is proportional to the square of . The rate constant is , so . At the start, , the concentration is . The student wants two values. The first is the concentration after . The second is the half-life, the time taken for the concentration to fall to half of its starting value. She also asks herself whether the half-life would be the same if the experiment started from a different concentration. Find the two values, and decide whether her guess is right.
Need a reminder? The key ideaShow
A separable equation becomes . Integrate both sides with one constant, then use the point.
The zeros of give constant solutions.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Given. with at .
Plan. Separate the variables, integrate, fix the constant at , then read off the two times.
Step 1. Separate and integrate. gives , so .
Step 2. The constant. At , . So .
Step 3. After . , so .
Step 4. The half-life. The concentration has fallen to half when . Then , so .
Answer: after , and a half-life of .
What if? With a start the half-life is . A start of gives . So her guess is right: for a second-order reaction the half-life depends on the starting concentration.
Check. The concentration falls from to in , and the fall is fastest at the start, as the squared term suggests.
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