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At elevated temperatures, methylisonitrile (CH3NC) isomerizes to acetonitrile (CH3CN): CH3NC (g) → CH3CN (g) At the start of the experiment, there are 0.200 mol of reactant (CH3NC) and 0 mol of product (CH3CN) in the reaction vessel. After 25 min of reaction, 0.108 mol of reactant (CH3NC) remain. The average rate of decomposition of methyl isonitrile, CH3NC, in this 25 min period is ________ mol/min.

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Answer:

Average rate of decomposition = 3.68 x 10⁻³ mol/min

Explanation:

CH₃NC (g) → CH₃CN (g)

Average rate of decomposition = [CH₃NC]/Δt

[CH₃NC] = initial [CH₃NC] - final [CH₃NC]

[CH₃NC] = 0.200 mol - 0.108 mol

[CH₃NC] = 0.092 mol

Average rate of decomposition = 0.092 mol / 25 min

Average rate of decomposition = 3.68 x 10⁻³ mol/min

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The average rate of decomposition of acetonitrile in this 25 min period is [tex]3.68 \times 10^-^3[/tex] mol/min.

What is isomerization?

The chemical reaction by which the compound is converted into its isomeric forms is called isomerization.

The equation is given as, methylisonitrile → acetonitrile

Calculate the moles of methylisonitrile as:

[methylisonitrile ]   = initial [methylisonitrile ] - final [methylisonitrile ]

[methylisonitrile] = 0.200 mol - 0.108 mol

[methylisonitrile] = 0.092 mol

The average rate of decomposition is calculated as:

[tex]\rm \dfrac{ [CH_3NC]}{\Delta t}[/tex]

[tex]\rm \dfrac{ [0.092 mol]}{25 min} = 3.68 \times 10^-3\;mol/min[/tex]

Thus, the average rate of decomposition of [tex]\rm [CH_3NC][/tex] is [tex]\rm 3.68 \times 10^-3\;mol/min[/tex]

Learn more about isomerization here:

https://brainly.com/question/2226351

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