(K_c) and also (K_p) space the equilibrium constants of gaseous mixtures. However, the difference in between the two constants is that (K_c) is defined by molar concentrations, conversely, (K_p) is identified by the partial pressure of the gasses inside a close up door system. The equilibrium constants do not incorporate the concentration of single components such as liquids and solidand castle may have units relying on the nature the the reaction (although thermodynamic equilibrium constants do not).

You are watching: How to convert kp to kc

Here room some easy measures on creating gas equilibrium constants (this is the exact same for recognize Kc, Kp, Ksp, Q and also etc.):.

How To write Gas Equilibrium Constants

In equilibrium equations, also though the arrows suggest both ways (( ightleftharpoons )) we usually associate the left together reactants and also the ideal as products. The assets are on the peak of the portion (the numerator). The reactants are on the BOTTOM that the fraction (the denominator). The concentrations of the products and reactants are always raised to the power of their coefficient in the well balanced invernessgangshow.netical equation.

The standard instance of composing Gas Equilibrium Constants are:

< aA + bB ; ightleftharpoons ; cC + dD onumber >

< K_c = dfrac^c^d^a^b onumber >

< K_p = dfrac(C)^c(D)^d (A)^a(B)^b onumber >

Example (PageIndex1):Thermal Decomposition of ( NH_4SH_(s) )

Consider the heat decomposition that ( NH_4SH_(s) ):

< NH_4SH_(s) ightleftharpoons NH_3 (g) + H_2S_(g) onumber >

This likewise is related to Ksp

< K_c = dfrac onumber >

but because (NH_4SH) is a solid, we get:

< K_c = dfrac<1> onumber >

< K_c = onumber >

As for Kp, the is the exact same as Kc, however instead of base < >, Kp supplies parentheses ( ):

< K_p = dfrac(NH_3)(H_2S)(NH_4SH) onumber >

< K_p = dfrac(NH_3)(H_2S)(1) onumber >

< K_p = (NH_3)(H_2S) onumber >

Example (PageIndex2):Hydrogen and Iodine

Consider the dual replacement reaction of hydrogen and iodine gas:

< H_2 (g) + I_2 (g) ightleftharpoons 2HI (g) onumber >

< K_c = dfrac^2 onumber >

< K_p = dfrac(HI)^2(H_2)(I_2) onumber >

## Definition of (K_c) and (K_p)

(K_c) is one equilibrium continuous in terms of molar concentrations and is usually characterized as:

in the basic reaction,

< aA + bB ightleftharpoons cC + dD onumber >

If a big (K_c) is created then over there are an ext products formed. Inversely, a small (K_c) indicates that the reaction favors the reactants.

(K_p) is one equilibrium continuous in terms of partial pressures. And also is usually identified as:

< K_p = dfrac(C)^c(D)^d(A)^a(B)^b onumber >

for the basic reaction

< aA + bB ightleftharpoons cC + dD onumber >

Homogeneous Equilibria: Reactants/Products all in a single phase. Because that example:

Heterogeneous Equilibria: Reactants/Products in much more than one phase. For example:

The relationship between the 2 equilibrium constants are:

< K_p = K_c (RT)^ Deltan onumber >

or

where,

( Delta n ) = (Total moles of gas top top the assets side) - (Total moles of gas on the reactants side). Thus ( Delta = (d + c) - (a + b) onumber >

## Relating Gas Equilibrium Constants to Equilibrium (K)

The value of K relies on whether the solution being calculated because that is utilizing concentrations or partial pressures. The gas equilibrium constants relate come the equilibrium (K) due to the fact that they room both derived from the appropriate gas law (PV = nRT).

(K_c) is the concentration that the reaction, it is usually shown as:

(K_p) is the quantity of partial press in the reaction, usually presented as:

< dfracp(C)p(D)p(A)p(B) onumber >

As we have actually seen above, Kp = Kc (RT)(Deltan),we deserve to derive this formula native the appropriate Gas Law.We understand that Kc is in terms Molarity (left(dfracMolesLiters ight)), and also we can likewise arrange the right Gas legislation (PV = nRT) as: (left(dfracnL ight) = left(dfracPRT ight)).

We know that Partial pressure is directly proportional to Concentration:

(P = left(dfracnL ight) * RT) Pressure have the right to be in devices of: Pascal (Pa), atmosphere (atm), or Torr.

Therefore we have the right to replace Kc through Molarity: the equation become, Kp = Kc (RT)(Deltan)

(RT)(Deltan) = (dfrac(RT)^c(RT)^d(RT)^a(RT)^b)

Also: (left(dfracnL ight) = left(dfracPRT ight)), deserve to be shown as Kc = Kp (RT)

K is likewise written the same as Kc and Kp:

( aA + bB ightleftharpoons cC + dD).

Example (PageIndex3)

< 2 NOBr_(g) ightleftharpoons 2 NO_(s) + Br_2 (g) onumber >

Given:

NOBr= 0.46 M NO= 0.1 M Br2 = 0.3M

To collection up Kc, the is (dfracProductsReactants)

< Kc= dfrac^2 ; ^2 onumber >

< Kc =dfrac<0.1>^2 ; <0.3><0.46>^2 onumber >

Kc= 0.0142 M

Example (PageIndex4)

N2O4 (l) is vital component that rocket fuel, in ~ 25 °C N2O4 is a colorless gas that partly dissociates right into NO2. The shade of one equilibrium mixture of this 2 gasses counts on their family member proportions, which room dependent on temperature. Equilibrium is created in the reaction ( N_2O_4 (g) ightleftharpoons 2NO_2 (g) ) at 25 °C.

Given:

3.00 together container 7.64 g N2O4 1.56 g NO2

What is the Kc for this reaction?

Solution

Step 1: transform grams to moles

mol N2O4 = 7.64 g * ( dfrac1 mol N_2 O_492.01 g ) = 8.303 * 10-2 mol

mol NO2 = 1.56 g * ( dfrac1 mol NO_246.01 g ) = 3.391 * 10-2 mol

Step 2: convert moles come Molarity (moles/L)

M = (dfrac8.303 * 10^-2 mol N_2O_43.00 L) = 0.0277 M

M = (dfrac3.391 * 10^-2 mol NO_23.00 L) = 0.0113 M

Step 3: write the Equilibrium consistent for Kc:

^2 = dfrac<0.0113>^2<0.0277> = 4.61 imes 10^-3 onumber >

Example (PageIndex5)

Calculate Kp for the reaction

Given:

N2= 0.79 moles O2 = 0.21 mole Temp = 2500 K

When equilibrium is established the mole percent that Nitrogen Oxide (NO) at 1.8%

1st step: produce an ice table:

ice cream

(N_2)

(O_2) (NO)
Initial 0.79 mol 0.21 mol 0 mol
Change -x -x +2x
Equilibrium (0.79 - x) (0.21 - x) (2x + 0)

2nd Step: uncover the Mole Percent (%) of products

XNO= 0.018

Xtotal = (0.79-x)+(.021-x)+(2x)= 1

XNO = (dfrac2xX_total)

0.018 = (dfrac2x1)

x = 0.009

3rd Step: collection up one equation

Kp = (dfracp(NO)^2p(N_2)p(O_2))

PV = NRT

(P = dfracdfrac^2V^2dfracn(N_2)(RT)Vdfracn(O_2)(RT)V)

(P = dfracn(NO)^2n(N_2)n(O_2))

4th Step: plug in values

(K_P = dfrac(2x)^2(0.79-x)(0.21-x))

x = 0.009

Kp = 2.1x10-3

## How the Gas Equilibrium Constants Relate to Reaction Quotient (Q)

The process of finding the Reaction Quotient (Qc) is the same as detect Kc and Kp, where the assets of the reaction is divided by the reaction of the reaction (left(dfracProductsReactants ight)) at any type of time not necessarily at equilibrium.

If a difficulty asks friend to uncover which means the reaction will shift in stimulate to attain equilibrium, and K is given, girlfriend would need to calculate for Q and also compare the two numbers.When compare K and Q:

K K > Q : since there are more reactants 보다 products, the reaction will produce more products to reach equilibrium, the reaction favors the products. K = Q : over there is no readjust in the products nor reactants, for this reason equilibrium is achieved.

A trick come remember come which what the reaction will certainly favor is:Put:K _ Q (in alphabet order! - or it will certainly not work)K The reaction will certainly favor the reactants due to the fact that reactants room on the left that the equation.K > Q : K ( ightarrow) QThe reaction will certainly favor the products due to the fact that products room on the right of the equation.K = Q : NO CHANGE

(See Relationship in between K and Q for much more information)

### Kc

1. Gaseous Hydrogen Iodide is inserted in a close up door container in ~ 425 (circ)C, whereby it partially decomposes come Hydrogen and Iodine:

2HI (g) ( ightleftharpoons) H2 (g) + I2 (g)

The following are given:

= 3.53 * 10-3 M