What is ${K_p}$? What goes on in a ${K_p}$equation?
Answer
590.1k+ views
Hint: We have to know that the forward and reverse reactions have the same rate when a reaction is in equilibrium. And as the forward and backward reactions continue, the concentrations of the reaction components remain stable at equilibrium. A chemical reaction's equilibrium constant is the value of its reaction quotient at chemical equilibrium, a condition reached by a dynamic chemical mechanism after a period of time has passed in which its structure shows no discernible propensity to alter.
Complete answer:
We need to remember that the ratio of concentrations at equilibrium for a reaction at a given temperature is described by equilibrium constants. In general, the letters $K$ or ${K_c}$ are used. Both concentrations are expressed in terms of molar concentration, as shown by the subscript c. When a reaction component is a gas, the sum of the chemical at equilibrium can also be expressed in terms of its partial pressure. When the equilibrium constant is expressed in terms of partial pressure, the expression for the equilibrium constant is ${K_p}$. Hence ${K_p}$ is the equilibrium constant expressed in terms of its partial pressure.
For a generic balanced gas-phase equation as given below,
\[aA\left( g \right) + bB\left( g \right) \rightleftharpoons cC\left( g \right) + dD\left( g \right)\]
In this equation, a moles of reactant A reacts with b moles of reactant B to give c moles of product C and d moles of product D.
If we know the partial pressures for each component at equilibrium, where the partial pressure of $A(g)$ is abbreviated as ${P_A}$, then
${K_p} = \dfrac{{{{({P_C})}^c}{{({P_D})}^d}}}{{{{({P_A})}^a}{{({P_B})}^b}}}$
This is the equation for ${K_p}$.
Note:
When we measure ${K_p}$, keep the following points in mind. The reaction must be balanced. Otherwise, the equilibrium constant's stoichiometric coefficients and exponents would be wrong. In the equilibrium term, pure liquids or solids have a concentration of $1$. This is the same as when you're figuring out ${K_c}$. ${K_p}$ is often written without the use of units. The units for all partial pressures used to calculate ${K_p}$ should be the same.
Complete answer:
We need to remember that the ratio of concentrations at equilibrium for a reaction at a given temperature is described by equilibrium constants. In general, the letters $K$ or ${K_c}$ are used. Both concentrations are expressed in terms of molar concentration, as shown by the subscript c. When a reaction component is a gas, the sum of the chemical at equilibrium can also be expressed in terms of its partial pressure. When the equilibrium constant is expressed in terms of partial pressure, the expression for the equilibrium constant is ${K_p}$. Hence ${K_p}$ is the equilibrium constant expressed in terms of its partial pressure.
For a generic balanced gas-phase equation as given below,
\[aA\left( g \right) + bB\left( g \right) \rightleftharpoons cC\left( g \right) + dD\left( g \right)\]
In this equation, a moles of reactant A reacts with b moles of reactant B to give c moles of product C and d moles of product D.
If we know the partial pressures for each component at equilibrium, where the partial pressure of $A(g)$ is abbreviated as ${P_A}$, then
${K_p} = \dfrac{{{{({P_C})}^c}{{({P_D})}^d}}}{{{{({P_A})}^a}{{({P_B})}^b}}}$
This is the equation for ${K_p}$.
Note:
When we measure ${K_p}$, keep the following points in mind. The reaction must be balanced. Otherwise, the equilibrium constant's stoichiometric coefficients and exponents would be wrong. In the equilibrium term, pure liquids or solids have a concentration of $1$. This is the same as when you're figuring out ${K_c}$. ${K_p}$ is often written without the use of units. The units for all partial pressures used to calculate ${K_p}$ should be the same.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

