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The rate constant for the second-order reaction: 2NOBr(g) → 2NO(g) + Br2(g) is 0.80/(M · s) at 10°C. Starting with a concentration of 0.86 M, calculate the concentration of NOBr after 95 s.

Respuesta :

Answer : The concentration of NOBr after 95 s is, 0.013 M

Explanation :

The integrated rate law equation for second order reaction follows:

[tex]k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)[/tex]

where,

k = rate constant = [tex]0.80M^{-1}s^{-1}[/tex]

t = time taken  = 95 s

[A] = concentration of substance after time 't' = ?

[tex][A]_o[/tex] = Initial concentration = 0.86 M

Now put all the given values in above equation, we get:

[tex]0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)[/tex]

[A] = 0.013 M

Hence, the concentration of NOBr after 95 s is, 0.013 M

okpalawalter8

The concentration of NOBr after 95 s is mathematically given as
NOBr=0.01129M

What is the concentration of NOBr after 95 s.?

Question Parameter(s):

Generally, the equation for the   is mathematically given as

[tex]\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\\\\frac{1}{[NOBr]} =\frac{0.8}{M*s}*95s+\frac{1}{0.080M}\\\\\frac{1}{[NOBr]}=\frac{29.3}{M}\\\\ NOBr =\frac{1}{29.2/M}[/tex]

NOBr=0.01129M

In conclusion, the concentration

NOBr=0.01129M

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