For the reaction 2NO(g) + Cl_(2)to2NOCl(g) the following data were collected. All the measurnments were taken at 263 K. a) Write the expression for rate law, b) Calculate the value of rate constant and specify its untis. c) What is the initial rate of disappearance of Cl_(2) in exp. 4?

For the reaction 2NO(g) + Cl_(2)to2NOCl(g) the following data were c

1.	For the reaction 2NO(g) + Cl_(2)to2NOCl(g) the following data were collected. All the measurnments were taken at 263 K. a) Write the expression for rate law, b) Calculate the value of rate constant and specify its untis. c) What is the initial rate of disappearance of Cl_(2) in exp. 4?
Answer» Solution :The RATE law equation may be expressed as: Rate=`k[A]^(p)[B]^(q)` Comparing experiments 1 and 2 `(Rate)_(1) = k[0.15]^(p)[0.15]^(q) = 0.60` `(Rate)_(2) = k[0.15]^(p)[0.30]^(q) = 1.20` DIVIDING eq. (ii) by eq. (i), `(Rate_(2))/(Rate_(1)) = (k[0.15]^(p)[0.30]^(q))/(k[0.15]^(p)[0.15]^(q))=1.20/0.60 = 2, [2]^(q) = [2]^(1)` or q=1 Comparing experiments 1 and 3 `(Rate_(1)) = k[0.15]^(p)[0.15]^(q)` `(Rate_(3)) = k[0.30]^(p)[0.15]^(q)` Dividing eq. (III) by eq (i) `(Rate_(3))/(Rate_(1)) = (k[0.30]^(p)[0.15]^(q))/(k[0.15]^(p)[0.15]^(q)) = 2.40/0.60 = 4,[2]^(p)=[2]^(2) or p=2` a) Expression for rate law, Rate = `k[NO]^(2)[Cl_(2)]` b) Rate constant (k) for the reaction may be calculated as: `k[0.15]^(p)[0.15]_(q) = 0.60, k0.150^(2)[0.15]^(1) = k60, k = p(0.60)/(0.003375) = 1.78 xx 10^(2)` since, the reaction is of third order : units of rate constant k =`L^(2)mol^(-2)min^(-1)` c) Initial rate of disappearancing of chlorine in experiment-4 Rate `= k[0.20mol L^(-1)]^(2)[0.25 mol L^(-1)] = (1.78 xx 10^(2)L^(2)min^(-1)) xx (0.25 mol L^(-1))^(3)` `= 2.78 mol L^(-1)min^(-1)`