For the chemical reaction A+2Bto2C+D. The experimentally determined information has been tabulated below: for the reaction, a)Calculate the order of reaction w.r.t. both the reactants A and B b) Write the expression for rate law. c) Calculate the value of the rate constant d) Write the expression for the rate of reaction in terms A and C.

For the chemical reaction A+2Bto2C+D. The experimentally determined i

1.	For the chemical reaction A+2Bto2C+D. The experimentally determined information has been tabulated below: for the reaction, a)Calculate the order of reaction w.r.t. both the reactants A and B b) Write the expression for rate law. c) Calculate the value of the rate constant d) Write the expression for the rate of reaction in terms A and C.
Answer» SOLUTION :a) Calculation of order with respect to A and B Let the rate law equation for the reaction be : rate = `k[A]^(x)[B]^(y)` The rates for the four experiments may be written as: 0.96 = `k[0.30]^(x)[0.30]^(y)` `0.384= k[0.60]^(x)[0.30]^(y)` `0.192 = k[0.30]^(x)[0.60]^(y)` `0.768 = k[0.60]^(x)[0.60]^(y)` Dividing eqn (ii) by eqn. (i), we get `(0.384)/(0.96) = (k[0.60]^(x)[0.30]^(y))/(k[0.30]^(x)[0.30]^(y))` or `4= (2)^(x)` `(2)^(2)=(2)^(x)` or x=2 Dividing eqn. (iii) by eqn. (i), we get `(0.192)/(0.096) = (k[0.30]^(x)[0.60]^(y))/(k[0.30]^(x)[0.30]^(y))or 4=(2)^(x)` `(2)^(2) = (2)^(x) or x=2`. Dividing Eqn. (iii)by eqn. (i), we get `0.192/0.096= (k[0.30]^(x)[0.60]^(y))/(k[0.30]^(x)[0.30]^(y))` or 2 = `(2)^(y)` `(2)^(') = (2)^(y)` or y=1 Order w.R.t. A= 2, B=1 ltrbgt Rate law expression , Rate(r) = `k[A]^(2)[B]^(1)` c) Rate constant(k) = `("Rate"(r))/([A]^(2)[B]^(2))=(0.096)/(0.027) = 3.56` d) Rate of reaction in TERMS A and C, Rate = `(-d[A])/(dt) = 1/2(d[C])/(dt)`