Answer: The concentration of ethene is 0.286 M
Explanation:
Rate law expression for first order kinetics is given by the equation:
[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]
where,
k = rate constant = [tex]2.25\times 10^{-2}s^{-1}[/tex]
t = time taken for decay process = 15.0 s
[tex][A_o][/tex] = initial amount of the reactant = 0.500 M
[A] = amount left after decay process = ?
Putting values in above equation, we get:
[tex]2.25\times 10^{-2}s^{-1}=\frac{2.303}{15.0s}\log\frac{0.500}{[A]}[/tex]
[tex][A]=0.357M[/tex]
The concentration of reactant consumed = (0.500 - 0.357) M = 0.143 M
For the given chemical reaction:
[tex]C_4H_8\rightarrow 2C_2H_4[/tex]
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
