Question

The optical rotation of an aqueous solution of a compound 'A' changes slowly due to slow fist order hydrolysis of ''A into 'B' and 'C' both of which are also optically active.specific rotations of A,B and C are +30degree, - 60degree and
+50degree respectively. starting with pure 'A', the time taken for rotation to change from +30degree to 0degree is 20 minutes. calculate t1/2 of the reaction?

Answer 1

the reaction is \(A \rightarrow B+C\)
at t= 0, the optical rotation is: +30

with the equation: 0.3x-0.6y+0.5x=0.3x-0.1y=optical rotation;

where y = [B]=[C] and x=[A],

i found (by trial and error): that 0=0.3(0.25)-0.1(0.75)

so the concentrations after 20 min were \([A_t]\)=0.25 and \([B_t]=[C_t]=0.75\)

Since it's a 1st order reaction: \(A_t=A_0*e^{(-\frac{ln(2)t}{t_{1/2}})} \rightarrow \large 0.25=1*e^{(-\frac{ln(2)*20min}{t_{1/2}})}\)

solve for half-life \(t_{1/2}\)

Answer 2

sorry the equation should read: " 0.3x-0.6y+0.5\(\color{red}y\)=0.3x-0.1y=optical rotation"

Answer 3

Can u tell me complete solution

Answer 4

you should try it first

Answer 5

what did you get?