i understand most of the problem but where did the 0.84 M come from?
a drug was injected into your body with a 2.00 M concentration. what would be its concentration after 0.010 seconds if k = 875 ^ -1
1st order reaction!
ln [ A ]t = - (875 ^ -1) (0.010 SEC) + ln (2.00)
= - 0.1769
[ A ] t = e^ -0.1769 = 0.84M
The concentration of 0.84 M came from solving the equation for the concentration of the drug after 0.010 seconds using the given information and the mathematical expression for a first-order reaction.
To understand how this was calculated, let's break it down step by step:
1. Given information:
- Initial concentration of the drug: 2.00 M
- Time elapsed: 0.010 seconds
- Rate constant (k) for the reaction: 875^(-1) (875 raised to the power of -1)
2. Calculate the natural logarithm of the concentration at time t:
- We use the equation for a first-order reaction: ln[A]t = -kt + ln[A]0
- [A]t represents the concentration at time t
- ln[A]t is the natural logarithm of [A]t
- kt is the product of the rate constant (k) and the time (t)
- ln[A]0 is the natural logarithm of the initial concentration [A]0
- Plugging in the given values: ln[A]t = -(875^(-1))(0.010 sec) + ln(2.00)
3. Simplifying the equation:
- -(875^(-1))(0.010 sec) = -0.1769 (rounded to four decimal places)
- The natural logarithm of 2.00 is approximately 0.6931 (rounded to four decimal places)
4. Calculate the concentration at time t:
- Using the equation ln[A]t = -kt + ln[A]0, plugging in the simplified values:
ln[A]t = -0.1769 + 0.6931
ln[A]t = 0.5162
- We can rewrite the equation as: [A]t = e^(ln[A]t), where e is the base of the natural logarithm (approximately 2.7183)
- [A]t = e^(0.5162)
- Calculating that gives us approximately 0.84 M (rounded to two decimal places)
So, the concentration of the drug after 0.010 seconds would be approximately 0.84 M.