ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: Studyinghard on July 10, 2010, 01:20:39 pm
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PCl3 + Cl2 -----> PCl5
2 mol of PCl5 is added to an empty 10 L vessel at 200 deg.
When equilibrium is reached there is 1.7mol PCl5 left. calculate the equilibrium constant at 200 deg
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k=[PCl3][CL2]/[PCl5]
[PCL5]=1.7/10= 0.17 M
[CL2]= 0.3mol/10L = 0.03M
[PCl3]= 0.03M
I wrote the k like that because since the reaction starts at the right side, the left side will give products
etc.
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k=[PCl3][CL2]/[PCl5]
[PCL5]=1.7/10= 0.17 M
[CL2]= 0.3mol/10L = 0.03M
[PCl3]= 0.03M
I wrote the k like that because since the reaction starts at the right side, the left side will give products
etc.
The working out for the concentrations is correct, however K should not be written that way. The product is supposed to be PCl5, as the equation is set out. I can see your reasoning, but you want to find the equilibrium constant for the equation that is specified.
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So i do 0.17/(0.03)2 = 188.89
?
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k=[PCl3][CL2]/[PCl5]
[PCL5]=1.7/10= 0.17 M
[CL2]= 0.3mol/10L = 0.03M
[PCl3]= 0.03M
I wrote the k like that because since the reaction starts at the right side, the left side will give products
etc.
The working out for the concentrations is correct, however K should not be written that way. The product is supposed to be PCl5, as the equation is set out. I can see your reasoning, but you want to find the equilibrium constant for the equation that is specified.
can't you k^-1 to find the reverse equation?
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k=[PCl3][CL2]/[PCl5]
[PCL5]=1.7/10= 0.17 M
[CL2]= 0.3mol/10L = 0.03M
[PCl3]= 0.03M
I wrote the k like that because since the reaction starts at the right side, the left side will give products
etc.
The working out for the concentrations is correct, however K should not be written that way. The product is supposed to be PCl5, as the equation is set out. I can see your reasoning, but you want to find the equilibrium constant for the equation that is specified.
can't you k^-1 to find the reverse equation?
Yep. But its not necessary if you can find K directly.
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So i do 0.17/(0.03)2 = 188.89
?
Yep, that looks right.
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In the reaction 2SO3(g) -----> 2SO2(g) + O2
0.10 mol SO3 is added to a 1L vessel. At equilibrium, there is 0.08 mol of the SO3 remainin. Calculate value of K
So will it be
K = (0.02)^2 x (0.01)
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0.08^2
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^^
yes, looks right :)
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Yeah, thats correct