For the system: I2(g) + Cl2(g) ↔ 2ICl(g) at equilibrium 0.310 moles of I2, 0.310 moles of Cl2 and 2.81 moles of ICl was found in a 1L vessel. 1 mole of ICl was then added to the vessel. Calculate the final equilibrium concentration of Cl2
Molarity of I2 = moles / volume ( in litres )
= 0.310 / 1
= 0.310 M
Molarity of Cl2 = 0.310 / 1 = 0.310
Molarity of ICl = 2.81/ 1 = 2.81
Kc = (ICl)^2 / ((I2) * (Cl))
= ( 2.81 )^2 / ( 0.310*0.310)
= 82.17
After adding 1 moles
ICl = 3.81
Now
I2 + Cl2 = 2 ICl
0.310 0.310 3.81 ( initially )
0.310+x 0.310+x 3.81-2x (equilibrium)
Kc = (ICl)^2 / ( I2*Cl2)
82.17 = (3.81-2x)^2 / ((0.310+x)*(0.310+x))
x = 0.0904
Cl2 = 0.310 + x = 0.310 + 0.0904 = 0.4004 M
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For the system: I2(g) + Cl2(g) ↔ 2ICl(g) at equilibrium 0.310 moles of I2, 0.310 moles...
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