What is the rate of reaction when c4h8 0.20 m

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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?

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Answer: The concentration of ethene is 0.286 M

Explanation:

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

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant = 2.25\times 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\times 10^{-2}s^{-1}=\frac{2.303}{15.0s}\log\frac{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


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