For which reaction is KP equal to KC?

For which reaction is KP equal to KC?

Kp only counts with gases molecule , while Kc only counts with aqueous solution+ gases. So here reactant and product both are gaseous that’s why only reaction having equal numbers of reactants and products will have kp=kc ie. change in number of moles is equal to 0.

Under what conditions for a reaction KP and KC are numerically equal?

Kp equals Kc when Δn = 0. This is true when the number of moles of gaseous products equals the number of moles of gaseous reactants in the balanced chemical equation. The value of Kp may also be less than Kc (for Δn < 0) or greater than Kc (for Δn > 0).

Is KC and KP the same?

Kc is the equilibrium constant when it is found through the use of concentrations, while Kp is the equilibrium constant when it is found through the use of partial pressures.

Under what condition KP KC and KX are equal?

S : Kc, Kp and Kx are the equilibrium constants of a reaction in terms of concentration, pressure and mole fraction respectively. E : Kc and Kp do not depend on equilibrium pressure but Kx depends upon equilibrium pressure if Δn=0.

Does KP depend on pressure?

Since the equilibrium constant Kp is a function of ΔGorxn which is defined for a specific composition (all reactants in their standard states and at unit pressure (or fugacity), changes in pressure have no effect on equilibrium constants for a fixed temperature.

Is KP dependent on temperature?

The position of equilibrium doesn’t need to move to keep Kp constant. Equilibrium constants are changed if you change the temperature of the system. Kc or Kp are constant at constant temperature, but they vary as the temperature changes. You can see that as the temperature increases, the value of Kp falls.

Does KP change with volume?

Because there is an equal number of moles on both sides of the reaction, an increase in volume will have no effect on the equilibrium and thus there is no shift in the direction. Similarly, when you decrease the volume there is no effect on the equilibrium.

Does KP have to be ATM?

of Kp. Since each term p(X)/p0(X) is a ratio of two pressures, it is a number. The logarithm function operates on numbers only; it does not operate on physical quantities. Therefore Kp must be a number, not some value with units such as atm – 1 .

What is unit of KC?

Kc is in terms of molarity and Kp is in terms of pressure. Also both of them are ratios of respective quantities [ ratio of molarity(s) in Kc and ratio of pressure(s) in Kp], so they should be dimensionless according to dimensional analysis. But in some places I have seen units mentioned along with both Kc and Kp.

Why are KC and KP taken as dimensionless?

Kc and Kp are also dimensionless, as they are defined properly using activities of the reactants and products which are dimensionless too. For simple calculations for Kp, dividing by the standard pressure of 1 bar for each component in the ratio brought to any power will always yield a dimensionless result.

Is KC affected by pressure?

Changes in pressure moves the system to one direction or the other (depending on the moles of reactant and product gases), but does not affect the value of kc or kp. The equilibrium is being desturbed by the pressure change and responds to re-establish the value of the equilibrium constant.

What is the relationship between Q and KC?

Q can be used to determine which direction a reaction will shift to reach equilibrium. If K > Q, a reaction will proceed forward, converting reactants into products. If K < Q, the reaction will proceed in the reverse direction, converting products into reactants. If Q = K then the system is already at equilibrium.

Does changing pressure affect equilibrium?

Le Chatelier’s principle implies that a pressure increase shifts an equilibrium to the side of the reaction with the fewer number of moles of gas, while a pressure decrease shifts an equilibrium to the side of the reaction with the greater number of moles of gas.

Why does pressure and concentration not affect KC?

If the gas is one of the reactants or products, this would affect one of the concentrations, and the reaction will have to shift to reestablish equilibrium. Again, this changes one of the rates, but does not affect the rate constants, so Kc is unaffected.

Why does a catalyst not change the equilibrium constant?

This is because a catalyst speeds up the forward and back reaction to the same extent and adding a catalyst does not affect the relative rates of the two reactions, it cannot affect the position of equilibrium. A catalyst speeds up the rate at which a reaction reaches dynamic equilibrium.

What is the value of KP for the reverse reaction?

1 Answer. Kc of reverse reaction should just be the reciprocal of forward Kc. So, Kc of reverse reaction is 11752.

What is the equilibrium constant for the reverse reaction?

The equilibrium expression written for a reaction written in the reverse direction is the reciprocal of the one for the forward reaction. K’ is the constant for the reverse reaction and K is that of the forward reaction.

What is the equation of equilibrium?

In order for a system to be in equilibrium, it must satisfy all three equations of equilibrium, Sum Fx = 0, Sum Fy = 0 and Sum M = 0. Begin with the sum of the forces equations. The simplest way to solve these force systems would be to break the diagonal forces into their component pars.