Question

In case of gaseous homogeneous reaction, the active mass of the reactant is given by the expression:
A. $\dfrac{{RT}}{P}$
B. $\dfrac{P}{{RT}}$
C. $\dfrac{{PV}}{{RT}}$
D. $\dfrac{{nRT}}{V}$

Answer 1

The reaction is a gaseous homogeneous reaction, so both the reactants and products are in the gaseous state. The active mass is also known as the molarity and is given by $\dfrac{n}{V}$, where n is the number of moles and V is the volume.

Complete step by step answer:
We know that the gas equation for an ideal gas is given as:
$PV = nRT$
Where P is the absolute pressure
V is the volume
n is the number of moles
R is the universal gas constant
And T is the absolute temperature.
The active mass is given by $\dfrac{n}{V}$or molarity, that is, the concentration of a reacting substance expressed usually in moles per litre. It can also be defined as the concentration of the portion of a dissolved electrolyte that is dissociated into ions and hence is capable of carrying the electric current.
Now the active mass is given by $\dfrac{n}{V}$. It can also be expressed as $\dfrac{w}{{\left( {MV} \right)}}$ where w is the given mass and M is the molar mass. $\dfrac{w}{M}$ gives the number of moles. The unit of active mass is moles/litre. It is usually represented by enclosing its formula in square brackets.
Now in the equation,
$PV = nRT$
$P = \dfrac{{nRT}}{V}$
$\dfrac{n}{v} = \dfrac{P}{{RT}}$
So, $\dfrac{n}{v}$ is equal to pressure by the product of universal gas constant and temperature.

Thus, the correct option is B.

Note:
$PV = nRT$ is applicable to ideal gas. Also, the active mass is directly proportional to the pressure and inversely proportional to the absolute temperature. Students must understand what active mass is and derive it from the gas law carefully. We use the gas law because the equation mentions that it is a gaseous homogeneous reaction.