SME
General Chemistry II · Problem type
Kp from Kc
Use this for “Kc = 0.50 for N₂ + 3H₂ ⇌ 2NH₃ at 400 K. Find Kp.”
Table
Temperature (T)
(K)
Equilibrium constant from pressures (Kp)
This is an example. Type your own values to start a new problem.
Kp
=
Kc
(R
T (K)
)
Formulas
Kp
=
Kc
×
(0.08206
×
T
)Δn
4.64 ×
10
-4
=
0.5
×
(0.08206 × 400)
-2
Assumptions
•
Kc uses concentrations in M and Kp uses partial pressures in atm, so R = 0.08206 L·atm/(mol·K).
•
Δn is the moles of gas on the product side minus the moles of gas on the reactant side, from the coefficients. Solids and liquids are left out.
•
When Δn = 0, (RT)⁰ = 1 and Kp = Kc at every temperature.
•
Temperatures are in kelvins (°C + 273.15).
Step-by-step
WE KNOW
Kc = 0.5, Δn = −2, T = 400 K
FIND
equilibrium constant from pressures (Kp)
Step 1 · Find equilibrium constant from pressures (Kp)
Use Kp = Kc × (0.08206 × T)Δn
Each gas concentration c becomes a pressure P = cRT. So K picks up one factor of RT for each mole of gas the reaction gains.
Kp
=
0.5
×
(0.08206 × 400)
-2
Kp
=
0.5
×
32.824
-2
Kp = 0.5 × 0.0009281
Kp
=
4.64 ×
10
-4
Check
Put the numbers back in. Both sides should match.
4.64 ×
10
-4
=
0.5
×
(0.08206 × 400)
-2
✓
Topic check
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