A 2.00-L container at equilibrium at 550 K contains gases at the following conce

Question

A 2.00-L container at equilibrium at 550 K contains gases at the following concentrations: 0.1420 M CH4, 0.1420 M CCl4, and 0.0380 M CH2Cl2. The relevant reaction is shown below. CH4(g) + CCl4(g) Á 2 CH2Cl2(g) Suppose that 1.530 g of CH2Cl2 is added to the 2-L container above, and the system is allowed to return to equilibrium. Find the final concentration of CH2Cl2 once equilibrium is restored. (The container remains at 550 K throughout.) Be sure to set up your work in an organized fashion so that you may receive some credit if you fail to solve the problem!

Explanation / Answer

V = 2 L

CH4 = 0.142

CCl4 = 0.142

CH2CL2 = 0.038

reaction = CH4 + CCl4 = 2CH2Cl2

m = 1.53 g of CH2Cl2 added to 2 L container...

find final concentration...

Initially...

K = (CH2Cl2)^2 /(CH4)(CCl4)

K = (0.038^2) / (0.142*10.42) = 0.00097591

then

Concentration of product added

(CH2Cl2) = mol/V = (mass/MW) / V = 1.53/(84.93) / 2 = 0.009

so

initially:

(CH2Cl2) = 0.009+0.038 - 2x

(CH4) = 0.142+x

(CCl4) = 0.142 +x

substitue in K

K = (CH2Cl2)^2 /(CH4)(CCl4)

0.00097591 = (0.047 - 2x)^2 / (0.142+x)^2

sqrt(0.00097591) = (0.047-2x) / (0.142+x)

0.03123(0.142 + x) = 0.047 - 2x

0.004434 + 0.03123x = 0.047 - 2x

0.004434-0.047 =( -2 -0.03123)x

x = (0.004434-0.047) / (( -2 -0.03123))

x = 0.02095

so

(CH2Cl2) = 0.009+0.038 - 2x = ( 0.009+0.038 - 2*0.02095) = 0.0051 M

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