Respuesta :
Answer: [tex][A_t]=3.875\ M.[/tex]
Explanation:
Given,
Initial concentration of HI=4.78 M.
Time taken, t=27.3 s.
Rate constant of reaction, [tex]k=1.80\times10^{-3}\ M^{-1}s^{-1}[/tex].
We know, concentration time equation for second order reaction is.
[tex]kt=ln\dfrac{[A_o]}{[A_t]}[/tex] ( here [tex]A_o\ and\ A_t[/tex] is initial concentration and concentration at time t respectively).
Putting all given values in above equation.
We get, [tex]1.80\times `10^{-3}\times 27.3=ln\dfrac{4.78}{[A_t]}[/tex]
[tex][A_t]=3.875\ M.[/tex]
Hence, this is the required solution.
Answer:
[tex]\frac{1}{A_t} = 3.875 M[/tex]
Explanation:
Given data:
Rate constant[tex] k = 1.80\times 10^{-3} M^{-1}ms^{-1}[/tex]
t = 27.3 s
initial concentration - 4.78 M
From Rate equation we have following relation
[tex]\frac{1}{A_t} = kt + \frac{1}{A_o}[/tex]
where
[tex]A_t[/tex] is concentration after time t
[tex]A_o[/tex] is initial concentration
plugging all value in above relation and solve for concentration at time t = 2.73 s
[tex]\frac{1}{A_t} = 1.80\times 10^{-3} \times 27.3 + \frac{1}{4.78}[/tex]
[tex]\frac{1}{A_t} = 3.875 M[/tex]
