How can you determine the rate of a reaction from a concentration-time graph?

Study for the WJEC GCSE Chemistry Exam. Utilize flashcards and multiple choice questions, each with hints and explanations. Prepare effectively for your exam!

Multiple Choice

How can you determine the rate of a reaction from a concentration-time graph?

Explanation:
The rate is how fast the concentration changes with time. On a concentration–time graph, that change is shown by the slope (gradient) of the curve at any moment. The steeper the slope, the faster the change. For a reactant, the concentration falls, so the slope is negative, but the rate is taken as the positive magnitude of that slope. So to determine the rate at a specific time, you read off the gradient at that time. If you want a numerical sense, you can imagine the concentration dropping from 0.60 M to 0.50 M over 2 s. The gradient would be (0.50 − 0.60) / 2 = −0.10 M/s, so the rate is 0.10 M/s. Final concentration and time to reach the end point don’t tell you how quickly the change occurred, and the area under the curve isn’t a measure of instantaneous rate for a reaction. The gradient at a chosen time directly reflects the instantaneous rate at that moment, which is why it’s the best answer.

The rate is how fast the concentration changes with time. On a concentration–time graph, that change is shown by the slope (gradient) of the curve at any moment. The steeper the slope, the faster the change. For a reactant, the concentration falls, so the slope is negative, but the rate is taken as the positive magnitude of that slope. So to determine the rate at a specific time, you read off the gradient at that time.

If you want a numerical sense, you can imagine the concentration dropping from 0.60 M to 0.50 M over 2 s. The gradient would be (0.50 − 0.60) / 2 = −0.10 M/s, so the rate is 0.10 M/s.

Final concentration and time to reach the end point don’t tell you how quickly the change occurred, and the area under the curve isn’t a measure of instantaneous rate for a reaction. The gradient at a chosen time directly reflects the instantaneous rate at that moment, which is why it’s the best answer.

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