Answer:
the original concentration of A = 0.0817092 M
Explanation:
A reaction is considered to be of first order it it strictly obeys the graphical equation method.
[tex]k_1 = \dfrac{2.303}{t}log \dfrac{a}{a-x}[/tex]
where;
k = the specific rate coefficient = 3.4 × 10⁻⁴ s⁻¹
t = time = 5.0 h = 5.0 × 3600 = 18000 seconds
a = initial concentration = ???
a - x = remaining concentration of initial concentration at time t = 0.00018 mol L⁻¹
[tex]3.4 \times 10^{-4}= \dfrac{2.303}{18000}log \dfrac{a}{0.00018}[/tex]
[tex]3.4 \times 10^{-4}= 1.27944 \times 10^{-4} \times log \dfrac{a}{0.00018}[/tex]
[tex]\dfrac{3.4 \times 10^{-4}}{1.27944 \times 10^{-4}}= log \dfrac{a}{0.00018}[/tex]
[tex]2.657= log \dfrac{a}{0.00018}[/tex]
[tex]10^{2.657}= \dfrac{a}{0.00018}[/tex]
[tex]453.94 = \dfrac{a}{0.00018}[/tex]
[tex]a =453.94 \times 0.00018[/tex]
a = 0.0817092 M
Thus , the original concentration of A = 0.0817092 M
