Answer:
[tex][CH_3CHO]=0.550M[/tex]
Explanation:
Hello.
In this case, since the rate law for this chemical reaction would be:
[tex]r=\frac{d[CH_3CHO]}{dt} =-k[CH_3CHO]^2[/tex]
The integrated rate law after its demonstration is:
[tex]\frac{1}{[CH_3CHO]} =kt+\frac{1}{[CH_3CHO]_0}[/tex]
Thus, for the given rate constant, initial concentration of CH3CHO and elapsed time, the resulting concentration is:
[tex]\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[/tex]
[tex]\frac{1}{[CH_3CHO]} =1.82L/mol[/tex]
[tex][CH_3CHO]=0.550M[/tex]
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