QJEEAdv04Ch0d78: JEE Advanced 2004 Chemistry - Class 12 JEEAdv04Ch0d78 Solution: For the given reactions, A + B → Products, following data were…

According to the given data, we see that when the concentration of A alone is doubled (0.1 $$\to$$ 0.2 mol L^$$-$$1), the rate of the reaction gets doubled (0.05 $$\to$$ 0.1 mol L^$$-$$1 s^$$-$$1). Therefore, the order with respect to A is 1.

When the concentration of B alone is doubled (0.1 $$\to$$ 0.2 mol L^$$-$$1), the rate of the reaction remains unaltered. Therefore, the order with respect to B is 0.

(1) The rate equation is, Rate = $$k[A][B{]}^0=k[A]$$

(2) Rate $$=k[A]$$ or, $$k=\frac{rate}{[A]}$$ or, $$k=\frac{0.05}{0.1}$$ $$\therefore$$ k = 0.5 s^$$-$$1