Question

For the reaction \[{{N}_{2}}{{O}_{5}}\xrightarrow{{}}2N{{O}_{2}}+\dfrac{1}{2}{{O}_{2}}\text{ }{{t}_{{\scriptscriptstyle 1\!/\!{ }_2}}}=24\text{ }hrs\] starting with \[10\text{ }g\] of \[{{N}_{2}}{{O}_{5}}\]how many grams of \[{{N}_{2}}{{O}_{5}}\] will remain after a period of \[96\] hours ?

Answer 1

Hint :We know that the half-life of a reaction is defined as the time required reducing the concentration of the reactant to half of its initial value. It is denoted by the symbol ${{t}_{1/2}}.$ The half-life of a reaction depends on the order of the reaction.

Complete Step By Step Answer:
The sum of the power of the concentration terms, on which the rate of reaction actually depends, as observed experimentally, is called the order of the reaction. In a first-order reaction rate varies as the first power of the concentration of the reactant, i.e. the rate increases as the number of times as the concentration of reactant is increased. For a first order reaction, the half-life is given by the equation:
For a 1st order reaction,
\[{{t}_{1/2}}=\text{ }\dfrac{0.693}{k}\left( k=rate\text{ }constant \right).\]
Therefore, \[k=\dfrac{0.693}{{{t}_{1/2}}}=\text{ }\dfrac{0.693}{24}=0.0289h{{r}^{-1}}\]
\[kt=\text{ }2.303\times log\left( \dfrac{a}{a-x} \right)~.......\left[ a=initial\text{ }amount,\text{ }a-x=amount\text{ }left \right]\]
Substituting the values in above equation;
\[0.0289\times 96\text{ }=\text{ }2.303\times log\left( \dfrac{10}{10-x} \right)\]
On further solving;
\[\Rightarrow \dfrac{10}{10-x}=antilog\left( 1.2 \right)\]
\[\therefore 10=158.5-15.85x\text{ }\Rightarrow 15.85x=148.5\]
Thus, \[x=9.36g\]
Now, just putting the values of \[x\] and then we \[a-x\]
$\therefore a-x=10-9.36=0.63g$
Therefore, $0.63$ grams of \[{{N}_{2}}{{O}_{5}}\] will remain after a period of \[96\] hours.

Note :
Remember that the half lifetime of a first-order reaction is independent of the reactant concentration. Thus, for one type of reaction if we have reactant concentration different in two sets of reactions, still both will show the same half-life time. The half lifetime of one type of first-order reaction will however be different for another type.