Question

At elevated temperatures, methylisonitrile (CH 3NC) isomerizes to acetonitrile (CH 3CN): CH 3NC (g) → CH 3CN (g) At the start of the experiment, there are 0.200 mol of reactant (CH 3NC) and 0 mol of product (CH 3CN) in the reaction vessel. After 25 min of reaction, 0.108 mol of reactant (CH 3NC) remain. The average rate of decomposition of methyl isonitrile, CH 3NC, in this 25 min period is __________ mol/min.

Answer 1

0.00368 mol / min

Step-by-step explanation:

The isomerization of methylisonitrile to acetonitrile is:

CH₃NC(g) → CH₃CN(g)

This reaction occurs at elevated temperatures.

If, in the beggining, you have 0.200mol of methylisonitrile and after 25 min 0.108mol, the moles of methylisonitrile that were descomposed are:

0.200mol - 0.108mol = 0.092mol.

In 25 min, the average rate of descomposition is:

0.092mol / 25min = 0.00368 mol / min