The decomposition of IBr(g)into I2 (g) and Br2 (g) is first order in IBr with k = 0.00255/sec.
(a)Starting with [IBr] = 1.50M, what will [IBr] become after 2.50 minutes?
(b)How long, in minutes, will it take for [IBr] to go from 0.500M to 0.100M?
(c)What is the half-life for this reaction in seconds?
(d)Enough IBr is added to an evacuated container to make [IBr] = 0.350M. How long will it take [I2] = 0.100M?
To answer these questions, we need to use the first-order reaction rate equation:
ln([A]/[A]₀) = -kt
Where:
[A] is the concentration of A at time t
[A]₀ is the initial concentration of A
k is the rate constant
t is the time
(a) Starting with [IBr] = 1.50 M, we want to find [IBr] after 2.50 minutes. We can use the rate equation as follows:
ln([IBr]/1.50) = -0.00255 * (2.50 * 60)
Solving for [IBr]:
[IBr] = 1.50 * e^(-0.00255 * (2.50 * 60))
(b) To find the time it takes for [IBr] to decrease from 0.500 M to 0.100 M, we can rearrange the rate equation and solve for time:
ln(0.100/0.500) = -0.00255 * t
Solving for t:
t = -ln(0.100/0.500)/0.00255
(c) The half-life of a first-order reaction can be determined using the equation:
t₁/₂ = ln(2)/k
where t₁/₂ is the half-life and k is the rate constant. In this case, we can use:
t₁/₂ = ln(2)/0.00255
(d) To find how long it will take for [I₂] to reach 0.100 M, we need to use the rate equation with [IBr] = 0.350 M. Rearranging the equation and solving for time:
ln(0.100/0) = -0.00255 * t
t = -ln(0.100/0)/0.00255
Note: ln(0) is not valid, so we consider it as 0 in this case, assuming [I₂] is initially 0.
Please note that these calculations involve assumptions and simplifications based on the first-order reaction kinetics. The actual reaction may have more complex rate laws.