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IF 2M ,1L SOLUTON OF ACETIC ACID IS ADDED TO 3M,1L ETHYL ALCOHOL THEN THE FOLLOWING ELEMENTARY REACTION TAKES PLACE CH3COOH+C2H5OH=CH3COOC2H5+H2O.IF EACH SOLUTION IS DILUTED BY 1LITER THEN THE INITIAL RATE BECOMES HOW MANY TIMES?A-4,B-2,C-0.25,D-0.5            SIR THE EQUAL SIGN I HAVE GIVEN ARE REVERSIBLE ARROWS.

Asked by amitjena226 16th November 2017, 8:25 PM
Answered by Expert
Answer:
because R a t e space equals K left square bracket C h subscript 3 C O O H right square bracket to the power of 1 left square bracket C subscript 2 H subscript 5 O H right square bracket to the power of 1
I n i t i a l space c o n c. space o f space a c i d space left parenthesis a right parenthesis equals space 2 M divided by L comma space c o n c. space o f space a l c o h o l left parenthesis b right parenthesis space equals space 3 M divided by L
space space space space space space therefore space r subscript 1 space equals space K space left square bracket a right square bracket to the power of 1 left square bracket b right square bracket to the power of 1
space space space space space space space space space space space space space space space space equals space K space left square bracket 2 right square bracket to the power of 1 left square bracket 3 right square bracket to the power of 1
space space space space space space space space space space space space r subscript 1 space equals 6 K 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 space space space space space space space space space space space space...................................... left parenthesis 1 right parenthesis
I f space w e space e a c h space r e a c tan t space i s space d i l u t e d space b y space 1 space l i t r e comma space i. e. comma space c o n c. space w i l l space b e c o m e space h a l f space
space space space space space space space space space space therefore space r subscript 2 space equals space K open square brackets a over 2 close square brackets to the power of 1 space space open square brackets b over 2 close square brackets to the power of 1
space space space space space space space space space space space space space space space space space space space equals space K space open square brackets 2 over 2 close square brackets to the power of 1 space open square brackets 3 over 2 close square brackets to the power of 1
space space space space space space space space space space space space space space space r subscript 2 space space end subscript equals space K open square brackets 3 over 2 close square brackets 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 space space space space space space space space space space space space space................................... left parenthesis 2 right parenthesis

B y space e q s. space left parenthesis 1 right parenthesis space a n d space left parenthesis 2 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 r subscript 1 over r subscript 2 equals fraction numerator 6 K over denominator open parentheses begin display style fraction numerator 3 K over denominator 2 end fraction end style close parentheses end fraction space equals 4
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 therefore space r subscript 1 equals 4 r subscript 2
Answered by Expert 17th November 2017, 6:12 PM
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