For a reaction with a rate constant of 0.5 s^-1, how long will it take for the concentration of a reactant to decrease to 25% of its initial value in a first-order reaction?
Practice Questions
1 question
Q1
For a reaction with a rate constant of 0.5 s^-1, how long will it take for the concentration of a reactant to decrease to 25% of its initial value in a first-order reaction?
1.386 seconds
2 seconds
4 seconds
8 seconds
For a first-order reaction, the time to reach 25% of the initial concentration is t = (ln(1/0.25))/k = (ln(4))/0.5 = 2.772/0.5 = 5.544 seconds.
Questions & Step-by-step Solutions
1 item
Q
Q: For a reaction with a rate constant of 0.5 s^-1, how long will it take for the concentration of a reactant to decrease to 25% of its initial value in a first-order reaction?
Solution: For a first-order reaction, the time to reach 25% of the initial concentration is t = (ln(1/0.25))/k = (ln(4))/0.5 = 2.772/0.5 = 5.544 seconds.
Steps: 8
Step 1: Identify the rate constant (k) given in the problem, which is 0.5 s^-1.
Step 2: Understand that for a first-order reaction, the formula to find the time (t) to reach a certain concentration is t = (ln(initial concentration / final concentration)) / k.
Step 3: Determine the initial concentration and the final concentration. The initial concentration is 100% and the final concentration is 25%.
Step 4: Substitute the values into the formula. Here, final concentration is 0.25 (which is 25% of 1). So, we have t = (ln(1 / 0.25)) / 0.5.
Step 5: Calculate 1 / 0.25, which equals 4. So now we have t = (ln(4)) / 0.5.
Step 6: Find the natural logarithm of 4, which is approximately 1.386.
Step 7: Divide 1.386 by 0.5 to find t. This gives t = 1.386 / 0.5 = 2.772 seconds.
Step 8: Therefore, the time it takes for the concentration to decrease to 25% of its initial value is approximately 2.772 seconds.
