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lutik1710 [3]
3 years ago
6

The following reaction was monitored as a function of time: A→B+C A plot of ln[A] versus time yields a straight line with slope

−4.3×10−3 /s. If the initial concentration of A is 0.260 M, what is the concentration after 225 s?
Chemistry
1 answer:
forsale [732]3 years ago
4 0

Answer : The concentration after 225 s is, 0.099 M

Explanation :

As we know that, the graph of ln [A] versus time yields a straight line with slope 'k'.

So, Slope = k = 4.3\times 10^{-3}s^{-1}

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 4.3\times 10^{-3}s^{-1}

t = time passed by the sample  = 225 s

a = initial amount of the reactant  = 0.260 M

a - x = amount left after decay process = ?

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

225=\frac{2.303}{4.3\times 10^{-3}}\log\frac{0.260}{a-x}

a-x=0.099M

Therefore, the concentration after 225 s is, 0.099 M

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