Answer: 0.011 M
Explanation:
[tex]C_3H_6\rightarrrow CH_2=CH-CH_3[/tex]
[tex]Rate=k[C_3H_6]^1[/tex]
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]6.0\times 10^{-4}\text{s}^{-1}[/tex]
t = time taken = 525 sec
a = initial amount of the reactant = 0.0226 mol/L
a - x = amount left = ?
Now put all the given values in above equation, we get
[tex]525=\frac{2.303}{6.0\times 10^{-4}}\log\frac{0.0226}{(a-x)}[/tex]
[tex](a-x)=0.011mol/L[/tex]
Thus the concentration after 525 s is 0.011 mol/L.
