Question

The reaction C4H8 → 2 C2H4 has a rate constant of 2.25×10–2 s –1 . What is [C2H4 ] after 15.0 s if the reaction container initially contained only C4H8 at an initial concentration of 0.500 M?

Answer 1

Answer: The concentration of ethene is 0.286 M

Step-by-step explanation:

Rate law expression for first order kinetics is given by the equation:

`k=(2.303)/(t)\log([A_o])/([A])`

where,

k = rate constant =
`2.25* 10^(-2)s^(-1)`

t = time taken for decay process = 15.0 s

`[A_o]` = initial amount of the reactant = 0.500 M

[A] = amount left after decay process = ?

Putting values in above equation, we get:

`2.25* 10^(-2)s^(-1)=(2.303)/(15.0s)\log(0.500)/([A])`

`[A]=0.357M`

The concentration of reactant consumed = (0.500 - 0.357) M = 0.143 M

For the given chemical reaction:

`C_4H_8\rightarrow 2C_2H_4`

1 mole of butene produces 2 moles of ethene

So, concentration of ethene = (2 × 0.143) M = 0.286 M

Hence, the concentration of ethene is 0.286 M