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Q.

A quantity of ethyl acetate is mixed with an excess of sodium hydroxide at 25°C. 100 c.c. of the mixture is immediately titrated against 0.05 N Hydrochloric acid, of which 75 c.c were required for neutralisation. After 30 minutes, 50 c.c. of the mixture required, similarly, 25 c.c. of the acid. When the original reaction of ester was complete, 25 c.c. of the mixture required 6.25 c.c. of the acid. Calculate the second order velocity constant (in mol/A/min) (at time =0 ) of the reaction, using concentration in moles per litre and time in minutes. Reaction is first order each w.r.t. NaOH and ester. Indicator chosen for above titration is such that, it gives end point when only hydrochloric acid reacts with (log 2=0.30log 3 = 0.48log 10 = 2.3) (Give your Answer after multiplying with a factor of 10 and excluding the decimal places)

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answer is 7.

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Detailed Solution

In the reaction we have,

RCOOR   +   OH    ROH  +  RCOO

    A                       B                  

t=0                     a                   -                      -

t=30               ax                                     x

t=                ab                                     b

a=0.0375; ax=0.025; x=0.0125

ab=0.0125

b=0.025

a×100=0.05×75

(ax)×50=0.05×25

(ab)×25=0.05×6.25

k×t=1(ab)ln b(bx)×(ax)a

k×30=1(ab)ln 0.0250.025×0.0250.0375

k=0.736 mol /l/min

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