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The rearrangement of methyl isonitrile 1CH₃NC₂ to acetonitrile 1CH₃CN₂ is a first-order reaction and has a rate constant of 5.11 × 10⁻⁵ s⁻¹ at 472 k. if the initial concentration of CH₃NC is 0.0340 m:

(a) What is the molarity of CH₃NC after 2.00 h?
(b) How many minutes does it take for the CH₃NC concentration to drop to 0.0300 m?
(c) How many minutes does it take for 20% of the CH₃NC to react?

1 Answer

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Final answer:

The rate constant is given for the rearrangement of methyl isonitrile to acetonitrile. The molarity of CH₃NC after 2.00 hours, the time for the concentration to drop to 0.0300 M, and the time for 20% of CH₃NC to react can be calculated using the first-order rate equation.

Step-by-step explanation:

The rate of the rearrangement of methyl isonitrile 1CH₃NC₂ to acetonitrile 1CH₃CN₂ is a first-order reaction. The rate constant is given as 5.11 × 10⁻⁵ s⁻¹ at 472 K. To answer the questions:

(a) To find the molarity of CH₃NC after 2.00 hours, we can use the first-order rate equation:

[CH₃NC]t = [CH₃NC]₀ * e^(-kt)

where [CH₃NC]t is the concentration at time t, [CH₃NC]₀ is the initial concentration, k is the rate constant, and t is the time elapsed. Plugging in the values, we get:

[CH₃NC]₂h = 0.0340 * e^(-5.11 × 10⁻⁵ * 2.00 * 3600)

(b) To find the time required for the CH₃NC concentration to drop to 0.0300 M, we can rearrange the first-order rate equation:

t = (-1/k) * ln([CH₃NC]t/[CH₃NC]₀)

Plugging in the values, we get:

t = (-1/5.11 × 10⁻⁵) * ln(0.0300/0.0340)

(c) To find the time required for 20% of CH₃NC to react, we can use the same first-order rate equation:

t = (-1/k) * ln([CH₃NC]t/[CH₃NC]₀)

Plugging in the values, we get:

t = (-1/5.11 × 10⁻⁵) * ln(0.80)

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