A 2.320 mol quantity of NOCl was initially placed in a 1.750 L reaction chamber at 400°C. After equilibrium was established, it was found that 28.70 percent of the NOCl has dissociated: 2NOCl (g) ⇆ 2NO (g) + Cl2 (g) Calculate the equilibrium constant Kc for the reaction.

Question

Chemistry

Ashtyn Cline

4 July, 02:37

A 2.320 mol quantity of NOCl was initially placed in a 1.750 L reaction chamber at 400°C. After equilibrium was established, it was found that 28.70 percent of the NOCl has dissociated: 2NOCl (g) ⇆ 2NO (g) + Cl2 (g) Calculate the equilibrium constant Kc for the reaction.

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Answers (1)

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Elizabeth Johns · Answer 1

Kc = 0.0307

Explanation:

Step 1: Data given

Number of moles NOCl = 2.320 moles

Volume = 1.750 L

Temperature = 400 °C

NOCl was 28.70 % dissociated

Step 2: The balanced equation

2NOCl (g) ⇔ 2NO (g) + Cl2 (g)

Step 3: Calculate molarity NOCl

[NOCl] = moles / volume

[NOCl] = 2.320 moles / 1.750 L

[NOCl] = 1.326 M

Step 4: Initial concentration

[NOCl] = 1.326 M

[NO] = 0M

[Cl2] = 0M

Step 5: Concentration at equilibrium

For 2 moles NOCl we'll have 2 moles NO and 1 mol Cl2

[NOCl] = (1.326 - 2X) M

[NO] = 2X M

[Cl2] = XM

Since NOCl is 28.70 % dissociated

([NOCl]0 - [NOCl]equi) = 0.2870[NOCl]0

2x = 0.2870 (1.326) ⇒ x = 0.190

Step 6: Calculate Kc

Kc = [NO]²[Cl2] / [NOCl]²

Kc = (2x) ²*x / (1.326 - 2x) ²

Kc = [2 (0.190) ]² * (0.190) / [1.326 - 2 (0.190) ]²

Kc = 0.0307