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solong [7]
2 years ago
7

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 concentrati

on of 0.86 M, calculate the concentration of NOBr after 95 s.
Chemistry
1 answer:
a_sh-v [17]2 years ago
4 0

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

Explanation :

The integrated rate law equation for second order reaction follows:

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

where,

k = rate constant = 0.80M^{-1}s^{-1}

t = time taken  = 95 s

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

[A]_o = Initial concentration = 0.86 M

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

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

[A] = 0.013 M

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

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Explanation:

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30.0L/473K = V2/294K

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Hey there,

Your answer would be

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