Question

The second-order reaction CH3CHO→CH4+CO has the reaction constant k of 6.73×10−5 L/mol s. If the initial concentration of CH3CHO is 0.551 mol/L, what is the concentration of CH3CHO after 45.0 seconds? Your answer should have 3 significant figures (three decimal places).

Answer 1

`[CH_3CHO]=0.550M`

Step-by-step explanation:

Hello.

In this case, since the rate law for this chemical reaction would be:

`r=(d[CH_3CHO])/(dt) =-k[CH_3CHO]^2`

The integrated rate law after its demonstration is:

`(1)/([CH_3CHO]) =kt+(1)/([CH_3CHO]_0)`

Thus, for the given rate constant, initial concentration of CH3CHO and elapsed time, the resulting concentration is:

`(1)/([CH_3CHO]) =6.73x10^(-5)(L)/(mol*s)*45.0s +(1)/(0.551mol/L)\\\\(1)/([CH_3CHO])=3.03x10^(-3)L/mol+1.81L/mol`

`(1)/([CH_3CHO]) =1.82L/mol`

`[CH_3CHO]=0.550M`

Best regards!