Answer:
Time required is 1.33 seconds
Explanation:
For first order reaction, the rate law expression is:
![kt = ln \frac{[A_{0}]}{[A_{t}]}](https://tex.z-dn.net/?f=kt%20%3D%20ln%20%5Cfrac%7B%5BA_%7B0%7D%5D%7D%7B%5BA_%7Bt%7D%5D%7D)
Where
A0 = initial concentration = 0.830 M
At = concentration after time t = 0.260 M
t = time in seconds = ?
k = rate constant = 0.870 s⁻¹
time = ](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7Bk%7D%5D%28ln%5B%5Cfrac%7B0.830%7D%7B0.260%7D%5D%29)
time = 
Answer:
I'm ok at it.
Explanation:
I can try to help if you need anything:)
Answer:
Answers are in the explanation.
Explanation:
- The half‑life of A increases as the initial concentration of A decreases. order: <em>2. </em>In the half-life of second-order reactions, the half-life is inversely proportional to initial concentration.
- A three‑fold increase in the initial concentration of A leads to a nine‑fold increase in the initial rate. order: <em>2. </em>The rate law of second-order is: rate = k[A]²
- A three‑fold increase in the initial concentration of A leads to a 1.73‑fold increase in the initial rate. order: <em>1/2. </em>The rate law for this reaction is: rate = k √[A]
- The time required for [A] to decrease from [A]₀ to [A]₀/2 is equal to the time required for [A] to decrease from [A]₀/2 to [A]₀/4. order: <em>1. </em>The concentration-time equation for first-order reaction is: ln[A] = ln[A]₀ - kt. That means the [A] decreasing logarithmically.
- The rate of decrease of [A] is a constant. order: <em>0. </em>The rate law is: rate = k -<em>where k is a constant-</em>