Topic 13 · page 4 of 12
Units of K
Kc and Kp have units that depend on the equation. Put the units into the expression and cancel, or count the coefficients. When everything cancels there are no units.
10 min read 13.3
Read first: Partial pressures and Kp , Heterogeneous systems
Put the units in and cancel 13.3
The spec asks for a value of K c K_\text{c} K c K p K_\text{p} K p
MethodFinding the units of K
Write the expression, with solids and pure liquids already left out.
Replace each [X] by mol dm⁻³ (for K c K_\text{c} K c p p p K p K_\text{p} K p
Cancel. Write the answer with positive and negative powers.
If everything cancels, write "no units".
Worked exampleUnits for the Haber process
N X 2 ( g ) + 3 H X 2 ( g ) ⇌ 2 N H X 3 ( g )
\ce{N2(g) + 3H2(g) <=> 2NH3(g)}
N X 2 ( g ) + 3 H X 2 ( g ) 2 NH X 3 ( g ) K c = [ N H X 3 ] 2 [ N X 2 ] [ H X 2 ] 3 ⟶ ( mol dm − 3 ) 2 ( mol dm − 3 ) ( mol dm − 3 ) 3 = ( mol dm − 3 ) − 2
K_\text{c} = \frac{[\ce{NH3}]^2}{[\ce{N2}]\,[\ce{H2}]^3}
\;\longrightarrow\;
\frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})\,(\text{mol dm}^{-3})^3}
= (\text{mol dm}^{-3})^{-2}
K c = [ N X 2 ] [ H X 2 ] 3 [ NH X 3 ] 2 ⟶ ( mol dm − 3 ) ( mol dm − 3 ) 3 ( mol dm − 3 ) 2 = ( mol dm − 3 ) − 2
Top: two concentrations, so ( mol dm − 3 ) 2 (\text{mol dm}^{-3})^2 ( mol dm − 3 ) 2 ( mol dm − 3 ) 4 (\text{mol dm}^{-3})^4 ( mol dm − 3 ) 4
The net power is 2 − 4 = −2, so the units are ( mol dm − 3 ) − 2 (\text{mol dm}^{-3})^{-2} ( mol dm − 3 ) − 2
Write that out: ( mol dm − 3 ) − 2 = mol − 2 dm 6 (\text{mol dm}^{-3})^{-2} = \text{mol}^{-2}\,\text{dm}^{6} ( mol dm − 3 ) − 2 = mol − 2 dm 6 K c K_\text{c} K c mol⁻² dm⁶ .
K p K_\text{p} K p atm 2 ÷ atm 4 \text{atm}^2 \div \text{atm}^4 atm 2 ÷ atm 4 atm⁻² .
A shortcut 13.3
The power of the unit is a count of gas or dissolved species in the equation.
For the Haber process: top 2, bottom 1 + 3 = 4, power = −2.
Equation
Top − bottom
K c K_\text{c} K c K p K_\text{p} K p
H X 2 ( g ) + I X 2 ( g ) ⇌ 2 H I ( g ) \ce{H2(g) + I2(g) <=> 2HI(g)} H X 2 ( g ) + I X 2 ( g ) 2 HI ( g ) 2 − 2 = 0
none
none
N X 2 ( g ) + 3 H X 2 ( g ) ⇌ 2 N H X 3 ( g ) \ce{N2(g) + 3H2(g) <=> 2NH3(g)} N X 2 ( g ) + 3 H X 2 ( g ) 2 NH X 3 ( g ) 2 − 4 = −2
mol⁻² dm⁶
atm⁻²
2 S O X 2 ( g ) + O X 2 ( g ) ⇌ 2 S O X 3 ( g ) \ce{2SO2(g) + O2(g) <=> 2SO3(g)} 2 SO X 2 ( g ) + O X 2 ( g ) 2 SO X 3 ( g ) 2 − 3 = −1
mol⁻¹ dm³
atm⁻¹
N X 2 O X 4 ( g ) ⇌ 2 N O X 2 ( g ) \ce{N2O4(g) <=> 2NO2(g)} N X 2 O X 4 ( g ) 2 NO X 2 ( g ) 2 − 1 = +1
mol dm⁻³
atm
P C l X 5 ( g ) ⇌ P C l X 3 ( g ) + C l X 2 ( g ) \ce{PCl5(g) <=> PCl3(g) + Cl2(g)} PCl X 5 ( g ) PCl X 3 ( g ) + Cl X 2 ( g ) 2 − 1 = +1
mol dm⁻³
atm
C a C O X 3 ( s ) ⇌ C a O ( s ) + C O X 2 ( g ) \ce{CaCO3(s) <=> CaO(s) + CO2(g)} CaCO X 3 ( s ) CaO ( s ) + CO X 2 ( g ) 1 − 0 = +1
mol dm⁻³
atm
C ( s ) + C O X 2 ( g ) ⇌ 2 C O ( g ) \ce{C(s) + CO2(g) <=> 2CO(g)} C ( s ) + CO X 2 ( g ) 2 CO ( g ) 2 − 1 = +1
mol dm⁻³
atm
3 F e ( s ) + 4 H X 2 O ( g ) ⇌ F e X 3 O X 4 ( s ) + 4 H X 2 ( g ) \ce{3Fe(s) + 4H2O(g) <=> Fe3O4(s) + 4H2(g)} 3 Fe ( s ) + 4 H X 2 O ( g ) Fe X 3 O X 4 ( s ) + 4 H X 2 ( g ) 4 − 4 = 0
none
none
C H X 3 C O O H ( l ) + C X 2 H X 5 O H ( l ) ⇌ C H X 3 C O O C X 2 H X 5 ( l ) + H X 2 O ( l ) \ce{CH3COOH(l) + C2H5OH(l) <=> CH3COOC2H5(l) + H2O(l)} CH X 3 COOH ( l ) + C X 2 H X 5 OH ( l ) CH X 3 COOC X 2 H X 5 ( l ) + H X 2 O ( l ) 2 − 2 = 0
none
not used
The last row has no K p K_\text{p} K p
Common mistake
Counting solids in the power, or flipping the sign. A power of −2 on the concentration unit is mol⁻² dm⁶, not mol² dm⁻⁶. Check by noticing that more on the bottom than the top always gives a negative power.
Exam technique
If the units cancel, write no units and show the cancelling if there is room. If a question gives pressures in kPa, quote the unit of K p K_\text{p} K p
Check yourself
Give the units of K c K_\text{c} K c K p K_\text{p} K p 2 N O ( g ) + O X 2 ( g ) ⇌ 2 N O X 2 ( g ) \ce{2NO(g) + O2(g) <=> 2NO2(g)} 2 NO ( g ) + O X 2 ( g ) 2 NO X 2 ( g )
Give the units of K c K_\text{c} K c K p K_\text{p} K p P C l X 5 ( g ) ⇌ P C l X 3 ( g ) + C l X 2 ( g ) \ce{PCl5(g) <=> PCl3(g) + Cl2(g)} PCl X 5 ( g ) PCl X 3 ( g ) + Cl X 2 ( g )
Explain why K c K_\text{c} K c H X 2 ( g ) + I X 2 ( g ) ⇌ 2 H I ( g ) \ce{H2(g) + I2(g) <=> 2HI(g)} H X 2 ( g ) + I X 2 ( g ) 2 HI ( g )
Write K c K_\text{c} K c 2 N O X 2 ( g ) + 7 H X 2 ( g ) ⇌ 2 N H X 3 ( g ) + 4 H X 2 O ( g ) \ce{2NO2(g) + 7H2(g) <=> 2NH3(g) + 4H2O(g)} 2 NO X 2 ( g ) + 7 H X 2 ( g ) 2 NH X 3 ( g ) + 4 H X 2 O ( g )
Give the units of K p K_\text{p} K p C a C O X 3 ( s ) ⇌ C a O ( s ) + C O X 2 ( g ) \ce{CaCO3(s) <=> CaO(s) + CO2(g)} CaCO X 3 ( s ) CaO ( s ) + CO X 2 ( g )
Give the units of K c K_\text{c} K c F e X 3 + ( a q ) + S C N X − ( a q ) ⇌ [ F e ( S C N ) ] X 2 + ( a q ) \ce{Fe^3+(aq) + SCN-(aq) <=> [Fe(SCN)]^2+(aq)} Fe X 3 + ( aq ) + SCN X − ( aq ) [ Fe ( SCN )] X 2 + ( aq )
Answers cover these until you have tried the questions
Top: 2. Bottom: 2 + 1 = 3. Power = −1. K c K_\text{c} K c mol⁻¹ dm³ . K p K_\text{p} K p atm⁻¹ .
Top: 1 + 1 = 2. Bottom: 1. Power = +1. K c K_\text{c} K c mol dm⁻³ . K p K_\text{p} K p atm .
K c = [ H I ] 2 [ H X 2 ] [ I X 2 ] K_\text{c} = \dfrac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]} K c = [ H X 2 ] [ I X 2 ] [ HI ] 2 ( mol dm − 3 ) 2 (\text{mol dm}^{-3})^2 ( mol dm − 3 ) 2 ( mol dm − 3 ) 2 (\text{mol dm}^{-3})^2 ( mol dm − 3 ) 2 no units .K c = [ N H X 3 ] 2 [ H X 2 O ] 4 [ N O X 2 ] 2 [ H X 2 ] 7 K_\text{c} = \dfrac{[\ce{NH3}]^2\,[\ce{H2O}]^4}{[\ce{NO2}]^2\,[\ce{H2}]^7} K c = [ NO X 2 ] 2 [ H X 2 ] 7 [ NH X 3 ] 2 [ H X 2 O ] 4 mol⁻³ dm⁹ .K p = p C O X 2 K_\text{p} = p_{\ce{CO2}} K p = p CO X 2 atm . Only one gas is in the expression, so the power is +1.K c = [ [ F e ( S C N ) ] X 2 + ] [ F e X 3 + ] [ S C N X − ] K_\text{c} = \dfrac{[\ce{[Fe(SCN)]^2+}]}{[\ce{Fe^3+}]\,[\ce{SCN-}]} K c = [ Fe X 3 + ] [ SCN X − ] [ [ Fe ( SCN )] X 2 + ] mol⁻¹ dm³ .