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NH2COONH4(s) ⇔ 2NH3(g) + CO2(g) In a closed vessel at 298K starting with only NH2COONH4(s) an equilibrium is established after sometimes. Then some ammonia is introduced and at the new equilibrium the partial pressure of NH3 equals the total pressure of old equilibrium. Find (Pt final)/(Pt initial) 
Asked by gargchahat2005 | 13 Dec, 2020, 05:16: PM
answered-by-expert Expert Answer
It is a question based on equilibrium.
N H subscript 4 C O O N H subscript 4 space left parenthesis s right parenthesis space left right double arrow space 2 N H subscript 3 left parenthesis g right parenthesis space plus space C O subscript 2 left parenthesis g right parenthesis
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 space space space space space space space space 2 p space space space space space space space space space space space space space space p

K subscript p equals open parentheses p subscript N H subscript 3 end subscript close parentheses squared cross times open parentheses p subscript C O subscript 2 end subscript close parentheses
space space space space space equals left parenthesis 2 P right parenthesis squared cross times left parenthesis P right parenthesis space space space space space space space space space space space............. left parenthesis 1 right parenthesis

N o w comma space i n space t h e space s e c o n d space c a s e comma space P subscript i left parenthesis i n i t i a l right parenthesis equals 3 P

N H subscript 4 C O O N H subscript 4 space left parenthesis s right parenthesis space left right double arrow space 2 N H subscript 3 left parenthesis g right parenthesis space plus space C O subscript 2 left parenthesis g right parenthesis
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 space space space space space space space space 3 P space space space space space space space space space space space space space space P to the power of apostrophe
K subscript p equals left parenthesis 3 P right parenthesis squared left parenthesis P apostrophe right parenthesis space space space space space space space space space space space............... left parenthesis 2 right parenthesis

F r o m space e q u a t i o n space left parenthesis 1 right parenthesis space a n d space left parenthesis 2 right parenthesis

left parenthesis 2 P right parenthesis squared left parenthesis P right parenthesis equals left parenthesis 3 P right parenthesis squared cross times left parenthesis P to the power of apostrophe right parenthesis
P to the power of apostrophe equals fraction numerator 4 P over denominator 9 end fraction

fraction numerator P subscript T subscript left parenthesis F i n a l right parenthesis end subscript over denominator P subscript T I left parenthesis I I n i t i a l end subscript right parenthesis end fraction equals fraction numerator 3 P plus P to the power of apostrophe over denominator 3 P end fraction equals fraction numerator 3 P plus begin display style fraction numerator 4 P over denominator 9 end fraction end style over denominator 3 P end fraction equals 31 over 27
Answered by Ravi | 14 Dec, 2020, 06:15: PM
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