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Question: At 400K, the half-life of a sample of a gaseous compound initially at 56.0 kPa is 340s. When the pre...

At 400K, the half-life of a sample of a gaseous compound initially at 56.0 kPa is 340s. When the pressure is 28.0 kPa, the half-life is 170s. The order of the reaction is:
A. 0
B. 2
C. 1
D. 12\dfrac{1}{2}

Explanation

Solution

Chemical reactions are assigned reaction orders that describe their kinetics. The order of the reaction is an experimental value that depends upon the number of the molecules in the rate-determining step. Further, the half-life of a species is the time it takes for the concentration of that substance to fall to half of its initial value.
Formula used: t12[A](1n){t_{\dfrac{1}{2}}}\infty \,{\left[ A \right]^{\left( {1 - n} \right)}} where, [A]\left[ A \right] is the concentration of the reactant and n is the order of the reaction.

Complete step by step answer:
For gaseous substances, instead of the concentration of the reaction, the pressure of the reactant is considered. Now, the relation between the order of the reaction and half-life is,

t12[A](1n) =t12[PA](1n)  \,\,\,\,\,{t_{\dfrac{1}{2}}}\infty \,{\left[ A \right]^{\left( {1 - n} \right)}} \\\ = {t_{\dfrac{1}{2}}}\infty \,{\left[ {{P_A}} \right]^{\left( {1 - n} \right)}} \\\

Where PA{P_A} is the pressure of the reactant. And n is the order f the reaction.
Now, from the given values, (t12)1=340s{\left( {{t_{\dfrac{1}{2}}}} \right)_1} = 340s , (PA)1=56.0 {\left( {{P_A}} \right)_1} = 56.0{\text{ }} and. (t12)2=170s{\left( {{t_{\dfrac{1}{2}}}} \right)_2} = 170s and (PA)2=28.0 {\left( {{P_A}} \right)_2} = 28.0{\text{ }} .therefore the equations are,

t12[PA](1n) 340[56.0](1n).....(1) {t_{\dfrac{1}{2}}}\infty \,{\left[ {{P_A}} \right]^{\left( {1 - n} \right)}} \\\ 340\infty {\left[ {56.0} \right]^{\left( {1 - n} \right)}}.....(1) \\\

And,

t12[PA](1n) 170[28.0](1n).....(2) {t_{\dfrac{1}{2}}}\infty \,{\left[ {{P_A}} \right]^{\left( {1 - n} \right)}} \\\ 170\infty {\left[ {28.0} \right]^{\left( {1 - n} \right)}}.....(2) \\\

Now, compare equation (1) and (2), and we get,

340170=[56.028.0](1n) 2=[2](1n) 1=(1n) 0=n  \dfrac{{340}}{{170}} = \,{\left[ {\dfrac{{56.0}}{{28.0}}} \right]^{\left( {1 - n} \right)}} \\\ 2 = \,{\left[ 2 \right]^{\left( {1 - n} \right)}} \\\ 1 = \,(1 - n) \\\ 0 = n \\\

Therefore, the order of the reaction is zero.
So, the correct option is A.

Additional information:
The definition of the rate of a reaction is the speed of a reaction by which the concentration of reactants decreases and the concentrations of products increase per unit time.
This rate of a reaction can be expressed in terms of concentration of reactants or products. The magnitude of the rate value does not depend upon the way of expression ( by reactant or product). For reactants, the rate shows as negative as throughout the reaction the concentration of the reactant decreases. On the other hand, the rate for the product shows as positive as the concentration of the product increases throughout the reaction.
For example, The reaction AA \to products. is a first-order reaction. The rate equation of the first-order reaction is r=k[A]r = k\left[ A \right] . Where the rate is r, the rate constant is k and [A]\left[ A \right] is the concentration of reactant A at a time t.

Note:
Remember that the rate of a reaction is the speed of a reaction by which the concentration of reactants decreases and the concentrations of products increase per unit time. The factors which affect the rate of a reaction is the concentration of reactants or product, temperature, the surface area of solid surface, pressure, and catalyst only. A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration