Question

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?

Answer 1

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

Answer 2

The concentration after 225 s is 0.099 M.

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

So, Slope = k =  $4.3*10^{-3}/s$

Rate law for first order kinetics:

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

where,

k = rate constant  =  $4.3*10^{-3}/s$

t = time passed by the sample  = 225 s

a = initial amount of the reactant  = 0.260 M

a - x = amount left after decay process = ?

On substituting the values:

$t=\frac{2.303}{k} log \frac{a}{a-x}\\\\t=\frac{2.303}{4.3*10^{-3}} log \frac{0.260}{a-x}\\\\a-x=0.099M$

Therefore, the concentration after 225 s is 0.099 M.

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