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CBSE Class 11-science Answered

Q) At 540K, 0.10 moles of PCL5 are heated in a 8 litre flask. The pressure of the equilibrium mixture is found to be 1.0 atm. Calculate Kp and Kc for the reaction.
Asked by Anish | 07 Feb, 2019, 05:20: PM
answered-by-expert Expert Answer
The reaction is 
 
 
space space space space space space space space space space space space space space space space space space space space space PCl subscript 5 space space rightwards harpoon over leftwards harpoon space PCl subscript 3 space end subscript space plus space Cl subscript 2 Initial space mole space space space space space space 0.1 space space space space space space space space space space 0 space space space space space space space space space space 0 Equ space mole space space space space space 0.1 minus straight alpha space space space space space space space straight alpha space space space space space space space space space space space straight alpha Total space no. space of space moles space equals space space space 0.1 minus straight alpha space plus space space straight alpha space space space plus space space straight alpha space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.1 space plus space straight alpha We space have comma straight V equals space 8 space litre straight P space equals space 1 space atm straight T space equals space 540 space straight K straight R equals space 0.082 PV space equals space nRT straight n equals space PV over RT open parentheses 0.1 plus straight alpha close parentheses space space space equals fraction numerator 1 cross times 8 over denominator 0.0821 cross times 540 end fraction space open parentheses 0.1 plus straight alpha close parentheses equals space 0.180 space space space space space space space space space space space space space straight alpha space equals space 0.180 minus 0.1 space space space space space space space space space space space space space space space space space equals space 0.08 space Now comma straight K subscript straight c space equals fraction numerator open square brackets open parentheses PCl subscript 3 close parentheses close square brackets open square brackets open parentheses Cl subscript 2 close parentheses close square brackets over denominator open square brackets open parentheses Pcl subscript 5 close parentheses close square brackets end fraction space space space space space space equals space fraction numerator 0.08 cross times 0.08 over denominator begin display style bevelled fraction numerator 0.1 minus straight alpha over denominator 8 end fraction end style end fraction space space space space space space space equals space fraction numerator 0.08 cross times 0.08 over denominator 8 open parentheses 0.1 minus 0.08 close parentheses end fraction space space space space space space equals fraction numerator 0.0064 over denominator 0.16 end fraction space straight K subscript straight c space equals space 0.04 space mol divided by lit straight K subscript straight p space equals space space straight K subscript straight c space cross times space open parentheses RT close parentheses to the power of increment straight n end exponent space space space space space equals space 0.04 cross times open parentheses 0.0821 cross times 540 close parentheses to the power of 1 space space space space space space equals 1.77 space atm
 
 
Equilibrium constant Kp 1.77 
 
                              Kc = 0.04
 
Answered by Varsha | 07 Feb, 2019, 07:09: PM

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