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sertanlavr [38]
3 years ago
7

The rate constant for a certain reaction is k = 5.40×10−3 s−1 . If the initial reactant concentration was 0.100 M, what will the

concentration be after 17.0 minutes?
Chemistry
1 answer:
IceJOKER [234]3 years ago
4 0

Answer : The concentration after 17.0 minutes will be, 4.05\times 10^{-4}M

Explanation :

The expression for first order reaction is:

[C_t]=[C_o]e^{-kt}

where,

[C_t] = concentration at time 't'  (final) = ?

[C_o] = concentration at time '0' (initial) = 0.100 M

k = rate constant = 5.40\times 10^{-3}s^{-1}

t = time = 17.0 min = 1020 s (1 min = 60 s)

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

[C_t]=(0.100)\times e^{-(5.40\times 10^{-3})\times (1020)}

[C_t]=4.05\times 10^{-4}M

Thus, the concentration after 17.0 minutes will be, 4.05\times 10^{-4}M

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What is the change in internal energy (ΔΕ) of a system when 5 kJ of work is done on the system while it releases 13 kJ of energy
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The change in internal energy (ΔΕ) of a system : -8 kJ

<h3>Further explanation  </h3>

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 \tt \Delta U=Q+W

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