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adelina 88 [10]
3 years ago
5

The first-order decomposition of cyclopropane has a rate constant of 6.7 * 10-4 5-1. If the initial concentration of cyclopropan

e 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
Chemistry
1 answer:
Sati [7]3 years ago
7 0

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:

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

Where,

[A_t] is the concentration at time t

So,

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

[A_t]=0.86 M

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Explanation:

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                     z = \frac{P \times V_{m}}{R \times T}

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So, calculate the molar volume as follows.

                V_{m} = \frac{z \times R \times T}{P}

                             = \frac{1.0005 \times 8.314 \times 10^{-3} m^{3}.kPa/mol K \times (60 + 273)K}{500 kPa}

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As molar mass of helium is 4 g/mol. Hence, calculate specific volume of helium as follows.

                    V_{sp} = \frac{V_{m}}{M_{w}}

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                           = 0.00139 m^{3}/g

                           = 0.00139 m^{3}/g \times \frac{1 g}{10^{-3}kg}

                                 = 1.39 m^{3}/kg

Thus, we can conclude that the specific volume of Helium in given conditions is 1.39 m^{3}/kg.

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