Second Order Kinetics?
I got the first part but I have tried over and over again to get the last two parts of question correct but cannot. Please help, thanks.
A reaction X ---> 2 Y proceeds according to a second-order rate law, with k = 0.246 L mol-1min-1.
If [X]0 = 0.524 mol/L at t = 0:
A) What is [X] at t = 45 min?
B) Assuming that the initial concentration of Y is zero ( [Y]0 = 0 ) and that there is no reverse
reaction, what is the concentration of Y at 45 min?
C) What is the rate of the reaction at t = 45 min?
Answers:
A) 0.077 mol L^-1
B) 0.89 mol L^-1
C) 1.4E-3 mol L^-1 min^-1
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Verified answer
A)
Use integrated 2nd order rate law.
(1/[X]) = k∙t + (1/[X]₀)
with
[X] reactant concentration at time t, [X]₀ initial reactant concentration, k rate constant.
Hence,
[X] = 1/(k∙t + (1/[X]₀))
= 1/( 0.246 L∙mol⁻¹∙min⁻¹ ∙ 45 min + ( 1 / 0.524 mol∙L⁻¹) )
= 0.077 mol∙L⁻¹
B)
According to reaction equation 2 moles of Y are formed per mole of X consumed. So the changes in
in concentration for x and y are related as:
∆[Y] = - 2∙∆[X]
When we consider the change form t=0 to t:
[Y] - [Y]₀ = - 2∙([X] - [X]₀)
with [Y]₀=0 follows for the product concentration at time t:
[Y] = 2∙([X]₀ - [X])
After 45 min
[Y] = 2∙(0.524 mol∙L⁻¹ - 0.077 mol∙L⁻¹)
= 0,894 mol∙L⁻¹
c)
2nd order rate law:
rate = k∙[X]²
After 45 min
rate = 0.246 L∙mol⁻¹∙min⁻¹ ∙ (0.077 mol∙L⁻¹)²
= 1.46×10⁻³ mol∙L⁻¹∙min⁻¹