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The first-order decomposition of cyclopropane has a rate constant of 6.7 * 10-4 5-1. If the initial concentration of cyclopropane is 1.33 M, what is the concentration of cyclopropane after 644 s? OA) 0.43 M B) 0.15 M C) 0.94 M D) 0.86 M

Respuesta :

Answer:

D) 0.86 M

Explanation:

Given that:

The rate constant, k = 6.7×10⁻⁴ s⁻¹

Initial concentration [A₀] = 1.33 M

Time = 644 s

Using integrated rate law for first order kinetics as:

[tex][A_t]=[A_0]e^{-kt}[/tex]

Where,

[tex][A_t][/tex] is the concentration at time t

So,

[tex][A_t]=1.33\times e^{-6.7\times 10^{-4}\times 644}[/tex]

[tex][A_t]=0.86 M[/tex]

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