Question

1. The equilibrium constant, K_c, for the following reaction is 42.2 at 288 K.

2CH₂Cl₂(g) <-----> CH₄(g) + CCl₄(g)  

When a sufficiently large sample of CH₂Cl₂(g) is introduced into an evacuated vessel at 288 K, the equilibrium concentration of CCl₄(g) is found to be 0.140 M.  

Calculate the concentration of CH₂Cl₂ in the equilibrium mixture. _____ M

2. A student ran the following reaction in the laboratory at 295 K:

2CH₂Cl₂(g) <-------> CH₄(g) + CCl₄(g)  

When she introduced 8.46×10^-2 moles of CH₂Cl₂(g) into a 1.00 liter container, she found the equilibrium concentration of CCl₄(g) to be 3.90×10^-2M.  

Calculate the equilibrium constant, K_c, she obtained for this reaction.  

K_c =

Answer 1

1. At equilibrium , [CH₄ ] = [CCl₄ ] = 0.140 M

and K_c = 42.2

For given reaction

K_c = [CH₄][CCl₄] / [CH₂Cl₂]²

42.2 . = (0.140)² / [CH₂Cl₂]²

[CH₂Cl₂] = 0.0216 M

2. At equilibrium , [CH₄ ] = [CCl₄ ] = 3.90 *10^-2 M = 0.039 M

Initial Moles of CH₂Cl₂ = 8.46 *10^-2 moles = 0.0846 moles

Volume = 1L

so, initial [CH₂Cl₂ ] = moles / volume = 0.0846 moles/ 1 L = 0.0846 M

[CH₂Cl₂] reacted = 2 * [CCl₄] = 2* 0.039 = 0.078 M

At equilibrium , [CH₂Cl₂] = Initial value - decomposed value = 0.0846 - 0.078 = 0.0066 M

Now, K_c = [CH₄][CCl₄] / [CH₂Cl₂]²

K_c = (0.039)² / (0.0066)² = 34.92