Answer:
(a)
[tex]0.0342M[/tex]
(b)
[tex]t_{1/2}=17.36s\\t_{1/2}=23.15s[/tex]
Explanation:
Hello,
(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:
[tex]\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\[/tex]
[tex][NOBr]=\frac{1}{29.2/M}=0.0342M[/tex]
(b) Now, for a second-order reaction, the half-life is computed as shown below:
[tex]t_{1/2}=\frac{1}{k[NOBr]_0}[/tex]
Therefore, for the given initial concentrations one obtains:
[tex]t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.072M}=17.36s\\t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.054M}=23.15s[/tex]
Best regards.
Explanation:
2NOBr(g) --> 2NO(g) 1 Br2(g)
Rate constant, k = 0.80
a) Initial concentration, Ao = 0.086 M
Final Concentration, A = ?
time = 22s
These parameters are connected with the equation given below;
1 / [A] = kt + 1 / [A]o
1 / [A] = 1 / 0.086 + (0.8 * 22)
1 / [A] = 11.628 + 17.6
1 / [A] = 29.228
[A] = 0.0342M
b) t1/2 = 1 / ([A]o * k)
when [NOBr]0 5 0.072 M
t1/2 = 1 / (0.072 * 0.80)
t1/2 = 1 / 0.0576 = 17.36 s
when [NOBr]0 5 0.054 M
t1/2 = 1 / (0.054 * 0.80)
t1/2 = 1 / 0.0432 = 23.15 s
