Chemistry: 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).

Bro no one can do this in time before it expires...

The second-order reaction CH3CHO→CH4+CO has the reaction

lilGi0

06.02.2022

[CH_3CHO]=0.550M

Explanation:

Hello.

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

$r=\frac{d[CH_3CHO]}{dt} =-k[CH_3CHO]^2$

The integrated rate law after its demonstration is:

$\frac{1}{[CH_3CHO]} =kt+\frac{1}{[CH_3CHO]_0}$

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

$\frac{1}{[CH_3CHO]} =6.73x10^{-5}\frac{L}{mol*s}*45.0s +\frac{1}{0.551mol/L}\\\\\frac{1}{[CH_3CHO]}=3.03x10^{-3}L/mol+1.81L/mol$

$\frac{1}{[CH_3CHO]} =1.82L/mol$

[CH_3CHO]=0.550M

Best regards!

0,0(0 оценок)