The equilibrium constant for the reaction Cl2(g)=2Cl(g) at 2000K is 0.55atm. A s
ID: 912568 • Letter: T
Question
The equilibrium constant for the reaction Cl2(g)=2Cl(g) at 2000K is 0.55atm. A sample of Cl2 is placed in a reaction bulb of volume 50.0mL, which is then heated to 2000K. The total pressure is found to be 1.36 atm. Calculate the mass of Cl2 originally placed in the bulb. The equilibrium constant for the reaction Cl2(g)=2Cl(g) at 2000K is 0.55atm. A sample of Cl2 is placed in a reaction bulb of volume 50.0mL, which is then heated to 2000K. The total pressure is found to be 1.36 atm. Calculate the mass of Cl2 originally placed in the bulb. The equilibrium constant for the reaction Cl2(g)=2Cl(g) at 2000K is 0.55atm. A sample of Cl2 is placed in a reaction bulb of volume 50.0mL, which is then heated to 2000K. The total pressure is found to be 1.36 atm. Calculate the mass of Cl2 originally placed in the bulb.Explanation / Answer
Solution:
Cl2 (g) <--> 2 Cl (g)
Kp = 0.55 atm
T = 2000 K
Total pressure = 1.36 atm
Volume = 50.0mL = 0.0050 mL
Now, total pressure of the system = pCl2 + pCl = 1.36 atm
(I)
Also, Kp for the above reaction = pCl2/pCl2 = 0.55 atm (II)
pCl2 = 1.36 atm – pCl (III)
Substituting (III) in (II), we get,
pCl2/(1.36 atm – pCl) = 0.55 atm
pCl2 + 0.55pCl – 0.748 = 0
Solving the polynomial equation for pCl, we get,
pCl = 0.6325 atm
thus, pCl2 = 1.36 – 0.6325 = 0.7274 atm
Now, assuming this as the ideal gas case, we can find number of moles of the Cl2 added initially.
pV = nRT
n = pV/RT = 0.7274*0.050/0.0821*2000
= 0.03637/164.2 = 2.21 * 10-4 moles
Now, n = mass of Cl2 gas/ Molar mass of Cl2 gas
Mass of Cl2 gas = n * Molar mass of Cl2 gas
= 2.21 * 10-4 * 70.906
= 0.01570 g
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