Question

Butadiene, c4h6 (used to make synthetic rubber and latex paints) dimerizes to c8h12 with a rate law of rate = 0.014 l/(mol·s) [c4h6]2. what will be the concentration of c4h6 after 3.0 hours if the initial concentration is 0.025 m?

Answer 1

`[\text{C}_4 \text{H}_6] \approx 5.2 * 10^(-3) \; \text{mol} \cdot \text{L}^(-1)`

Step-by-step explanation:

From the rate law:

`\text{d} [\text{C}_4 \text{H}_6] /\text{d}t = - 0.014 \cdot [\text{C}_4 \text{H}_6]^(2)`

where

• 
  `[\text{C}_4 \text{H}_6]` the reactant concentration, in
  `\text{mol}\cdot \text{L}^(-1)` (or equivalently,
  `\text{M}`
• 
  `t` the time into the reaction process, in seconds, and
• 
  `\text{d} [\text{C}_4 \text{H}_6] /\text{d}t` the reaction rate in differential form.

Note the negative sign in front of the right hand side of the equation. Rearrange the rate law expression to separate the two variables,

• concentration
  `[\text{C}_4 \text{H}_6]` and
• time
  `t`.

`-[\text{C}_4 \text{H}_6]^(-2) \cdot \text{d} [\text{C}_4 \text{H}_6] = 0.014 \cdot \text{d}t`

Implicitly integrate both sides of the expression (with the power rule) to obtain the general expression (i.e., the one with the arbitary constant "
`C`") for concentration
`[\text{C}_4 \text{H}_6]` at given time
`t`:

`\int(- [\text{C}_4 \text{H}_6]^(-2) )\cdot \text{d} [\text{C}_4 \text{H}_6] = \int 0.014 \cdot \text{d}t`

`[\text{C}_4 \text{H}_6]^(-1) = 0.014 \cdot \text{t} + C`

`[\text{C}_4 \text{H}_6] = 1/ (0.014 \cdot \text{t} + C)`

Now, solve for the value of
`C` for this particular configuration (i.e., with an initial,
`t = 0` concentration
`[\text{C}_4 \text{H}_6] = 0.025 \text{mol}\cdot \text{L}^(-1)`, as seen in the question)

`1/ (0.014 \cdot \text{t} + C) = 1\; /\; C = [\text{C}_4 \text{H}_6]_{\text{initial}} = 0.025`

`C = 1/0.025 = 40.0`

Thus, the
`\text{C}_4\text{H}_6` concentration at time
`t` given an initial
`\text{C}_4\text{H}_6` concentration of
`0.025 \; \text{mol}\cdot \text{L}^(-1)` would be

`[\text{C}_4 \text{H}_6](t) = 1/(0.014 \cdot t +40)`

where again,
`t` the time into the reaction in seconds.

The question is asking for the concentration at
`t = 3 \; \text{hrs} = 3 * 3600 \; \text{seconds}`

`[\text{C}_4\text{H}_6 ](10800) = 5.2 * 10^(-3) \; \text{mol} \cdot \text{L}^(-1)`

Reference:

Lannah Lua, Ciara Murphy, and Victoria Blanchard, "Second-Order Reactions," Libretext Chemistry