For first degree equation , time for 75% completion of reaction is X times half time of that reaction , where X is ?
1 Answer
Explanation:
As you know, the rate of a first-order reaction depends linearly on the concentration of a single reactant. The rate of a first-order reaction that takes the form
color(blue)(A -> "products")
can thus be written as
color(blue)("rate" = - (d["A"])/dt = k * ["A"])" " , where
Now, in Integral form (I won't go into the derivation here), the rate law for a first-order reaction is equal to
color(blue)(ln( (["A"])/(["A"_0])) = - k * t)" " , where
Now, the half-life of the reaction is the time needed for the concentration of the reactant to reach half of its initial value.
You can say that
t = t_"1/2" implies ["A"] = 1/2 * ["A"_0]
Plug this into the equation for the rate law to get
ln( (1/2 * color(red)(cancel(color(black)(["A"_0]))))/(color(red)(cancel(color(black)(["A"_0]))))) = -k * t_"1/2"
This means that you have
t_"1/2" = ln(1/2)/(-k) = (ln(1) - ln(2))/(-k) = ln(2)/k
In your case, you want to figure out how many half0lives must pass in order for
t = t_"x" implies ["A"] = 25/100 * ["A"_0] = 1/4 * ["A"_0]
Once again, plug this into the rate law equation to get
ln( (1/4 * color(red)(cancel(color(black)(["A"_0]))))/(color(red)(cancel(color(black)(["A"_0]))))) = - k * t_"x"
This means that you have
t_"x" = ln(1/4)/(-k) = (ln(1) - ln(4))/(-k) = ln(4)/k
But since
ln(4) = ln(2^2) = 2 * ln(2)
you can say that
t_"x" = 2 * overbrace(ln(2)/k)^(color(red)(t_text(1/2))) = color(green)(2 * t_"1/2")
Therefore,
X = 2
Simply put, the reaction will be