Equilibrium Partial Pressure of Substance C Calculator

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Chemical equilibrium

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Equilibrium Constant

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Relation between Equilibrium Constant and Degree of Dissociation

Relation between Vapour Density and Degree of Dissociation

Relationship between Different Equilibrium Constants

Thermodynamics in Chemical Equilibrium

✖Equilibrium constant for partial pressure is the value of its reaction quotient at chemical equilibrium with respect to partial pressure.ⓘ Equilibrium Constant for Partial Pressure [K_p]			+10% -10%
✖Equilibrium Partial Pressure A is the partial pressure for both homogeneous and heterogeneous reactions involving gas A.ⓘ Equilibrium Partial Pressure A [P_A]			+10% -10%
✖Number of Moles of A is the no. of moles of reactant A present in the equilibrium mixture.ⓘ Number of Moles of A [a]			+10% -10%
✖Equilibrium Partial Pressure B is the partial pressure for both homogeneous and heterogeneous reactions involving gas B.ⓘ Equilibrium Partial Pressure B [p_B]			+10% -10%
✖No. of Moles of B is the no. of moles of reactant B present in the equilibrium mixture.ⓘ No. of Moles of B [b]			+10% -10%
✖Equilibrium Partial Pressure D is the partial pressure for both homogeneous and heterogeneous reactions involving gas D.ⓘ Equilibrium Partial Pressure D [p_D]			+10% -10%
✖No. of Moles of D is the no. of moles of product D present in the equilibrium mixture.ⓘ No. of Moles of D [d]			+10% -10%
✖No. of Moles of C is the no. of moles of product C present in the equilibrium mixture.ⓘ No. of Moles of C [c]			+10% -10%

✖Equilibrium Partial Pressure C is the partial pressure for both homogeneous and heterogeneous reactions involving gas C.ⓘ Equilibrium Partial Pressure of Substance C [p_C]

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Formula

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Equilibrium Partial Pressure of Substance C

Formula

`"p"_{"C"} = (("K"_{"p"}*("P"_{"A"}^"a")*("p"_{"B"}^"b"))/("p"_{"D"}^"d"))^(1/"c")`

Example

`"80.0228Bar"=(("150mol/L"*(("0.77Bar")^"17")*(("50Bar")^"3"))/(("40Bar")^"7"))^(1/"9")`

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Equilibrium Partial Pressure of Substance C Solution

STEP 0: Pre-Calculation Summary

Formula Used

Equilibrium Partial Pressure C = ((Equilibrium Constant for Partial Pressure*(Equilibrium Partial Pressure A^Number of Moles of A)*(Equilibrium Partial Pressure B^No. of Moles of B))/(Equilibrium Partial Pressure D^No. of Moles of D))^(1/No. of Moles of C)
p_C = ((K_p*(P_A^a)*(p_B^b))/(p_D^d))^(1/c)
This formula uses 9 Variables

Variables Used

Equilibrium Partial Pressure C - (Measured in Pascal) - Equilibrium Partial Pressure C is the partial pressure for both homogeneous and heterogeneous reactions involving gas C.
Equilibrium Constant for Partial Pressure - (Measured in Mole per Cubic Meter) - Equilibrium constant for partial pressure is the value of its reaction quotient at chemical equilibrium with respect to partial pressure.
Equilibrium Partial Pressure A - (Measured in Pascal) - Equilibrium Partial Pressure A is the partial pressure for both homogeneous and heterogeneous reactions involving gas A.
Number of Moles of A - Number of Moles of A is the no. of moles of reactant A present in the equilibrium mixture.
Equilibrium Partial Pressure B - (Measured in Pascal) - Equilibrium Partial Pressure B is the partial pressure for both homogeneous and heterogeneous reactions involving gas B.
No. of Moles of B - No. of Moles of B is the no. of moles of reactant B present in the equilibrium mixture.
Equilibrium Partial Pressure D - (Measured in Pascal) - Equilibrium Partial Pressure D is the partial pressure for both homogeneous and heterogeneous reactions involving gas D.
No. of Moles of D - No. of Moles of D is the no. of moles of product D present in the equilibrium mixture.
No. of Moles of C - No. of Moles of C is the no. of moles of product C present in the equilibrium mixture.

STEP 1: Convert Input(s) to Base Unit

Equilibrium Constant for Partial Pressure: 150 Mole per Liter --> 150000 Mole per Cubic Meter (Check conversion here)
Equilibrium Partial Pressure A: 0.77 Bar --> 77000 Pascal (Check conversion here)
Number of Moles of A: 17 --> No Conversion Required
Equilibrium Partial Pressure B: 50 Bar --> 5000000 Pascal (Check conversion here)
No. of Moles of B: 3 --> No Conversion Required
Equilibrium Partial Pressure D: 40 Bar --> 4000000 Pascal (Check conversion here)
No. of Moles of D: 7 --> No Conversion Required
No. of Moles of C: 9 --> No Conversion Required

STEP 2: Evaluate Formula

Substituting Input Values in Formula

p_C = ((K_p*(P_A^a)*(p_B^b))/(p_D^d))^(1/c) --> ((150000*(77000^17)*(5000000^3))/(4000000^7))^(1/9)

Evaluating ... ...

p_C = 8002279.94090706

STEP 3: Convert Result to Output's Unit

8002279.94090706 Pascal -->80.0227994090706 Bar (Check conversion here)

FINAL ANSWER

80.0227994090706 ≈ 80.0228 Bar <-- Equilibrium Partial Pressure C

(Calculation completed in 00.004 seconds)

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National Institute of Information Technology (NIIT), Neemrana

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< 21 Properties of Equilibrium Constant Calculators

Equilibrium Partial Pressure of Substance A

Go Equilibrium Partial Pressure A = (((Equilibrium Partial Pressure C^No. of Moles of C)*(Equilibrium Partial Pressure D^No. of Moles of D))/(Equilibrium Constant for Partial Pressure*(Equilibrium Partial Pressure B^No. of Moles of B)))^(1/Number of Moles of A)

Equilibrium Partial Pressure of Substance B

Go Equilibrium Partial Pressure B = (((Equilibrium Partial Pressure C^No. of Moles of C)*(Equilibrium Partial Pressure D^No. of Moles of D))/(Equilibrium Constant for Partial Pressure*(Equilibrium Partial Pressure A^Number of Moles of A)))^(1/No. of Moles of B)

Equilibrium Partial Pressure of Substance C

Go Equilibrium Partial Pressure C = ((Equilibrium Constant for Partial Pressure*(Equilibrium Partial Pressure A^Number of Moles of A)*(Equilibrium Partial Pressure B^No. of Moles of B))/(Equilibrium Partial Pressure D^No. of Moles of D))^(1/No. of Moles of C)

Equilibrium Partial Pressure of Substance D

Go Equilibrium Partial Pressure D = ((Equilibrium Constant for Partial Pressure*(Equilibrium Partial Pressure A^Number of Moles of A)*(Equilibrium Partial Pressure B^No. of Moles of B))/(Equilibrium Partial Pressure C^No. of Moles of C))^(1/No. of Moles of D)

Equilibrium Constant with respect to Partial Pressure

Go Equilibrium Constant for Partial Pressure = ((Equilibrium Partial Pressure C^No. of Moles of C)*(Equilibrium Partial Pressure D^No. of Moles of D))/((Equilibrium Partial Pressure A^Number of Moles of A)*(Equilibrium Partial Pressure B^No. of Moles of B))

Equilibrium Mole Fraction of Substance A

Go Equilibrium Mole Fraction A = (((Equilibrium Mole Fraction C^No. of Moles of C)*(Equilibrium Mole Fraction D^No. of Moles of D))/(Equilibrium Constant for Mole Fraction*(Equilibrium Mole Fraction B^No. of Moles of B)))^(1/Number of Moles of A)

Equilibrium Mole Fraction of Substance B

Go Equilibrium Mole Fraction B = (((Equilibrium Mole Fraction C^No. of Moles of C)*(Equilibrium Mole Fraction D^No. of Moles of D))/(Equilibrium Constant for Mole Fraction*(Equilibrium Mole Fraction A^Number of Moles of A)))^(1/No. of Moles of B)

Equilibrium Constant for Reverse Reaction

Go Reverse Equilibrium Constant = ((Equilibrium Concentration of A^Number of Moles of A)*(Equilibrium Concentration of B^No. of Moles of B))/((Equilibrium Concentration of C^No. of Moles of C)*(Equilibrium Concentration of D^No. of Moles of D))

Equilibrium Mole Fraction of Substance C

Go Equilibrium Mole Fraction C = ((Equilibrium Constant for Mole Fraction*(Equilibrium Mole Fraction A^Number of Moles of A)*(Equilibrium Mole Fraction B^No. of Moles of B))/(Equilibrium Mole Fraction D^No. of Moles of D))^(1/No. of Moles of C)

Equilibrium Mole Fraction of Substance D

Go Equilibrium Mole Fraction D = ((Equilibrium Constant for Mole Fraction*(Equilibrium Mole Fraction A^Number of Moles of A)*(Equilibrium Mole Fraction B^No. of Moles of B))/(Equilibrium Mole Fraction C^No. of Moles of C))^(1/No. of Moles of D)

Equilibrium Constant with respect to Mole Fraction

Go Equilibrium Constant for Mole Fraction = ((Equilibrium Mole Fraction C^No. of Moles of C)*(Equilibrium Mole Fraction D^No. of Moles of D))/((Equilibrium Mole Fraction A^Number of Moles of A)*(Equilibrium Mole Fraction B^No. of Moles of B))

Molar Concentration of Substance A

Go Concentration of A = (((Concentration of C^No. of Moles of C)*(Concentration of D^No. of Moles of D))/(Reaction Quotient*(Concentration of B^No. of Moles of B)))^(1/Number of Moles of A)

Molar Concentration of Substance B

Go Concentration of B = (((Concentration of C^No. of Moles of C)*(Concentration of D^No. of Moles of D))/(Reaction Quotient*(Concentration of A^Number of Moles of A)))^(1/No. of Moles of B)

Molar Concentration of Substance C

Go Concentration of C = ((Reaction Quotient*(Concentration of A^Number of Moles of A)*(Concentration of B^No. of Moles of B))/(Concentration of D^No. of Moles of D))^(1/No. of Moles of C)

Molar Concentration of Substance D

Go Concentration of D = ((Reaction Quotient*(Concentration of A^Number of Moles of A)*(Concentration of B^No. of Moles of B))/(Concentration of C^No. of Moles of C))^(1/No. of Moles of D)

Reaction Quotient

Go Reaction Quotient = ((Concentration of C^No. of Moles of C)*(Concentration of D^No. of Moles of D))/((Concentration of A^Number of Moles of A)*(Concentration of B^No. of Moles of B))

Equilibrium Constant for Reversed Reaction when Multiplied with Integer

Go Equilibrium Constant Multiplied = 1/(Equilibrium Constant^Number)

Equilibrium Constant for Reaction when Multiplied with Integer

Go Equilibrium Constant Multiplied = (Equilibrium Constant^Number)

Weight of Reactant given Active Mass

Go Weight of Solute = Active mass*Molecular Weight

Active Mass

Go Active mass = Weight of Solute/Molecular Weight

Equilibrium Constant for Reverse Reaction given Constant for Forward Reaction

Go Reverse Equilibrium Constant = 1/Equilibrium Constant

Equilibrium Partial Pressure of Substance C Formula

What is equilibrium constant with respect to partial pressure?

Kp, equilibrium constant with respect to partial pressure has exactly the same format as Kc, irrespective of the amounts of A, B, C and D, except that partial pressures are used instead of concentrations. The gases on the right-hand side of the chemical equation are at the top of the expression, and those on the left at the bottom.

How to Calculate Equilibrium Partial Pressure of Substance C?

Equilibrium Partial Pressure of Substance C calculator uses Equilibrium Partial Pressure C = ((Equilibrium Constant for Partial Pressure*(Equilibrium Partial Pressure A^Number of Moles of A)*(Equilibrium Partial Pressure B^No. of Moles of B))/(Equilibrium Partial Pressure D^No. of Moles of D))^(1/No. of Moles of C) to calculate the Equilibrium Partial Pressure C, The Equilibrium partial pressure of substance C formula is defined as the partial pressure for both homogeneous and heterogeneous reactions involving gas C. Equilibrium Partial Pressure C is denoted by p_C symbol.

How to calculate Equilibrium Partial Pressure of Substance C using this online calculator? To use this online calculator for Equilibrium Partial Pressure of Substance C, enter Equilibrium Constant for Partial Pressure (K_p), Equilibrium Partial Pressure A (P_A), Number of Moles of A (a), Equilibrium Partial Pressure B (p_B), No. of Moles of B (b), Equilibrium Partial Pressure D (p_D), No. of Moles of D (d) & No. of Moles of C (c) and hit the calculate button. Here is how the Equilibrium Partial Pressure of Substance C calculation can be explained with given input values -> 0.10151 = ((150000*(77000^17)*(5000000^3))/(4000000^7))^(1/9).

FAQ

What is Equilibrium Partial Pressure of Substance C?

The Equilibrium partial pressure of substance C formula is defined as the partial pressure for both homogeneous and heterogeneous reactions involving gas C and is represented as p_C = ((K_p*(P_A^a)*(p_B^b))/(p_D^d))^(1/c) or Equilibrium Partial Pressure C = ((Equilibrium Constant for Partial Pressure*(Equilibrium Partial Pressure A^Number of Moles of A)*(Equilibrium Partial Pressure B^No. of Moles of B))/(Equilibrium Partial Pressure D^No. of Moles of D))^(1/No. of Moles of C). Equilibrium constant for partial pressure is the value of its reaction quotient at chemical equilibrium with respect to partial pressure, Equilibrium Partial Pressure A is the partial pressure for both homogeneous and heterogeneous reactions involving gas A, Number of Moles of A is the no. of moles of reactant A present in the equilibrium mixture, Equilibrium Partial Pressure B is the partial pressure for both homogeneous and heterogeneous reactions involving gas B, No. of Moles of B is the no. of moles of reactant B present in the equilibrium mixture, Equilibrium Partial Pressure D is the partial pressure for both homogeneous and heterogeneous reactions involving gas D, No. of Moles of D is the no. of moles of product D present in the equilibrium mixture & No. of Moles of C is the no. of moles of product C present in the equilibrium mixture.

How to calculate Equilibrium Partial Pressure of Substance C?

The Equilibrium partial pressure of substance C formula is defined as the partial pressure for both homogeneous and heterogeneous reactions involving gas C is calculated using Equilibrium Partial Pressure C = ((Equilibrium Constant for Partial Pressure*(Equilibrium Partial Pressure A^Number of Moles of A)*(Equilibrium Partial Pressure B^No. of Moles of B))/(Equilibrium Partial Pressure D^No. of Moles of D))^(1/No. of Moles of C). To calculate Equilibrium Partial Pressure of Substance C, you need Equilibrium Constant for Partial Pressure (K_p), Equilibrium Partial Pressure A (P_A), Number of Moles of A (a), Equilibrium Partial Pressure B (p_B), No. of Moles of B (b), Equilibrium Partial Pressure D (p_D), No. of Moles of D (d) & No. of Moles of C (c). With our tool, you need to enter the respective value for Equilibrium Constant for Partial Pressure, Equilibrium Partial Pressure A, Number of Moles of A, Equilibrium Partial Pressure B, No. of Moles of B, Equilibrium Partial Pressure D, No. of Moles of D & No. of Moles of C and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.

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