(b) (i) KF crystallizes in fcc structure like sodium chloride. Calculate the distance between K^(+) and F^(-) in KF . (given: density of KF is 2.48g cm^(-3)) (ii) How do concentration of the reactant influence the rate of reaction?

(b) (i) KF crystallizes in fcc structure like sodium chloride. Calcula

1.	(b) (i) KF crystallizes in fcc structure like sodium chloride. Calculate the distance between K^(+) and F^(-) in KF . (given: density of KF is 2.48g cm^(-3)) (ii) How do concentration of the reactant influence the rate of reaction?
Answer» Solution :(b) (i) Density of KF `2.48 g cm^(-3) " " rho = (nM)/(a^(3)N_(A))` n=4 (for FCC) `rho = (4 xx 58)/(a^(3) xx 6.023 xx 10^(23))` `2.48 = (4 xx 58)/(a^(3) xx 6.023 xx 10^(23))` `a^(3) = (4 xx 58)/(2.48 xx 6.023 xx 10^(23))` `a^(3) = (232)/(14.93 xx 10^(23))` `a^(3) = 15.54 xx 10^(-23)` a= 537.5 pm `d= (a)/(sqrt2)` (For fcc) `d= (537.5)/(1.414)` d= 380.12 pm (ii) The rate of a reaction increases with the increase in the concentration of the reactants. The effect of concentration is explained on the basis of collision theory of reaction rates. According to this theory, the rate of a reaction depends upon the number of collisions between the reacting molecules. Higher the concentration, GREATER is the possiblity for collision and hence the rate.