Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: heen20 on July 24, 2005, 01:27:30 PM
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when a 3.00 mol dm-3 sample of PCL5 is heated to 250 degrees the decomposition
PCL5 (g) <------> PCL3 (g) + CL2 (g)
occurs and reaches equilibrium with Kc = 1.80. What is the composition of the eqilibirum mixture? what percentage of PCL5 has decomposed at 250 degrees??
can any1 help me with this question??
i have no idea of how to even start this problem!!!!
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Calculate unknown concentrations at equilibrium and solve a quadratic equatuion.
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PCL5 (g) <------> PCL3 (g) + CL2 (g)
3-x ... x . x
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when a 3.00 mol dm-3 sample of PCL5 is heated to 250 degrees the decomposition
PCL5 (g) <------> PCL3 (g) + CL2 (g)
occurs and reaches equilibrium with Kc = 1.80. What is the composition of the eqilibirum mixture? what percentage of PCL5 has decomposed at 250 degrees??
can any1 help me with this question??
i have no idea of how to even start this problem!!!!
the same question i answered on another forum. Here's the answer:
Kc = [PCl3][Cl2] / [PCl5]
we start with 3,00 M PCl5 and due to the decomposition x mol PCl5 reacted and resulted into the forming of x mol Cl2 and PCl3. So we get the following equation (hence all molecules are in a same volume so concentration can be replaced by mols):
1,8 = x^2 / (3 - x) which we need to solve for x ( mol reacted PCl5 and thus mols Cl2 and PCl3)
--> 5,4 - 1,8x = x^2 --> x^2 + 1,8x - 5,4 = 0
now we get the values x as x(1) = (-1,8 + sqrt 24,84) / 2 = 1,591987159
and x(2) = (-1,8 - sqrt 24,84) / 2 = -3,391987159 which does not satisfie, because there can't be formed any PCl5.
That's why 1,591987159 mol PCl5 reacted resulting in the following mols molecules in equilibrium:
1,308012841 mol PCl5 and 1,591987159 mol Cl2 and PCl3. That means that 1,591987159 /3 * 100 = 53,1 % PCl5 is decomposed.