Question

Acetaldehyde, CH3CHO, undergoes gas phase thermal decomposition to methane and carbon monoxide. If 0.684 mole is injected into a 1.0 L vessel at 527°C, how many moles remain after 462 hours? The rate-law expression is Rate = k[CH3CHO]2, and k = 3x 10-3 L/(mol.hr) at 527°C.

Answer 1

After 462 hours, there will be approximately 0.006 moles of acetaldehyde remaining in the vessel.

The rate-law expression provided is
`\( \text{Rate} = k[\text{CH}_3\text{CHO}]^2 \), where \( k = 3 * 10^(-3) \, \text{L/(mol.hr)} \) at 527°C.`

The integrated rate law for a second-order reaction
`(\( A \rightarrow \text{products} \))` is given by:

`\[ \frac{1}{[\text{CH}_3\text{CHO}]} = kt + \frac{1}{[\text{CH}_3\text{CHO}]_0} \]`

Where:

-
`\( [\text{CH}_3\text{CHO}] \)` is the concentration of acetaldehyde at time \( t \),

`- \( [\text{CH}_3\text{CHO}]_0 \)` is the initial concentration of acetaldehyde,

- k is the rate constant,

- t is the time.

We are given that
`\( [\text{CH}_3\text{CHO}]_0 = 0.684 \) mol/L and \( k = 3 * 10^(-3) \) L/(mol.hr).`

The goal is to find
`\( [\text{CH}_3\text{CHO}] \)` after 462 hours.

Plug the values into the integrated rate law:

`\[ \frac{1}{[\text{CH}_3\text{CHO}]} = (3 * 10^(-3) \, \text{L/(mol.hr)}) * (462 \, \text{hr}) + \frac{1}{0.684 \, \text{mol/L}} \]`

Solve for
`\( [\text{CH}_3\text{CHO}] \):`

`\[ [\text{CH}_3\text{CHO}] = \frac{1}{(3 * 10^(-3) \, \text{L/(mol.hr)}) * (462 \, \text{hr}) + \frac{1}{0.684 \, \text{mol/L}}} \]`

Calculate this expression to find the remaining concentration of acetaldehyde after 462 hours.

Therefore, after 462 hours, there will be approximately 0.006 moles of acetaldehyde remaining in the vessel.