How do you move the 1 from 1/[A] over to the other side of the equation without

Question

How do you move the 1 from 1/[A] over to the other side of the equation without having the value for [A]?

The Integrated Rate Law for a Second-Order Reaction The reaction of butadiene gas (Calle) with itself produces Caliz gas as follows: 2C4H6(g) C8H12(g) The reaction is second order with a rate constant equal to 5.76 x 10-2 L/mol/min under certain conditions. If the initial concentration of butadiene is 0.200 M, what is the concentration remaining after 10.0 min? Solution We use the integrated form of the rate law to answer questions regarding time. For a second-order reaction, we have: We know three variables in this equation: [Alo = 0.200 mol/L, k = 5.76 min. Therefore, we can solve for [A], the fourth variable: 10-2 L/mol/min, and t 10.0 = (5.76x10-2 L mol-i min-1) (10 min) + 0.200 mol-1 = (5.76 x 10-1 L mol-1)+5.00 L mol-1 = 5.58 Lnol-1 [A] 1.79x10-1 mol L-1 Therefore 0.179 mol/L of butadiene remain at the end of 10.0 min, compared to the 0.200 mol/L that was originally present. Check Your Learning If the initial concentration of butadiene is 0.0200 M, what is the concentration remaining after 20.0 min? Answer: 0.0196 mol/L The integrated rate law for our second-order reactions has the form of the equation of a straight line:

Explanation / Answer

[A0] = 0.0200 M

[A] =?

t = 20 min

Rate constant k = 5.76*10^-2 L/mol/min

1/[A] = kt + 1/[A0]

=(5.76*10^-2 * 20) + 1/(0.0200)

1/[A] = (51.152/1)

By Cross multiplication

[A] = 0.0196 mol/L

For second order reaction

1/[A] = kt + 1/[A0]

rearrange the equation

1/[A] =( [A0] kt + 1)/[A0]

[A0] =( [A0] kt + 1)*[A]

Or

[A] = [A0] / ( [A0] kt + 1)

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