Question

Consider the following reaction.
N2(g) + O2(g) 2 NO(g)
If the equilibrium partial pressures of N2, O2, and NO are 0.15 atm, 0.33 atm, and 0.050 atm, respectively, at 2200°C, what is KP?

Answer 1

Concepts and reason

The concept used to solve this problem is determined the Equilibrium constant.

Fundamentals

At equilibrium, the rate of forward reaction and the rate of reverse reaction; both are same, and will not change with time, than the value of the reaction quotient is called equilibrium constant.

The equilibrium constant indicate by the sign of . 

At equilibrium, the value of the reaction quotient is called equilibrium constant.

is a ratio of the equilibrium concentrations of the product to the equilibrium concentration of the reactants.

For example the equilibrium constant is calculated as follows:

The equilibrium constant ()

$\begin{array}{l}\\{K_{{\rm{eq}}}} = \frac{{{\rm{[Products]}}}}{{[{\rm{Reactants}}]}}\\\\{\rm{ }} = {\rm{ }}\frac{{{{\left[ {{\rm{ C}}} \right]}^c}{{\left[ {{\rm{ D}}} \right]}^d}}}{{{{\left[ {\rm{A}} \right]}^a}{{\left[ {\rm{B}} \right]}^b}}}\\\end{array}$

In the above reaction; a, b, c and d are coefficients

The equilibrium constant can be expressed in terms of partial pressures of the reactants and products.

According to the ideal gas law:

Here, is the pressure, is the volume, is the gas constant, is the temperature and is the number of moles. is the concentration.

Therefore; the equilibrium constant; for gaseous reaction;

The given balance reaction is as follows:

The equilibrium constant; for gaseous reaction;

According to the problem the equilibrium partial pressures of = 0.15 atm, partial pressures of = 0.33 atm, and partial pressures of = 0.050 atm, at

The equilibrium constant () is calculated as follows:

$\begin{array}{l}\\{\rm{ }}{K_{\rm{p}}} = {\rm{ }}\frac{{{P_{{\rm{NO}}}}^2}}{{{P_{{{\rm{N}}_{\rm{2}}}}}^1\;{P_{{{\rm{O}}_{\rm{2}}}}}^1}}\\\\{\rm{ }} = {\rm{ }}\frac{{{{\left[ {{\rm{0}}{\rm{.050}}} \right]}^2}}}{{{{\left[ {{\rm{0}}{\rm{.15}}} \right]}^1}{{\left[ {{\rm{0}}{\rm{.33}}} \right]}^1}}}\\\\{\rm{ }} = {\rm{ }}\frac{{{\rm{0}}{\rm{.0025}}}}{{0.0495}}\\\\{\rm{ }} = {\rm{ 0}}{\rm{.0505}}\\\end{array}$

Ans:

Thus, for given reaction is .