H2 + I2 ==> 2HI
rate = k(H2)(I2)
you know k, you can calculate (H2) and (I2) from PV = nRT
For part b just triple the (H2) calculated in part a and recalculate rate
When 0.13 g of H2 and 0.18 g of I2 are confined to a 150. mL reaction vessel and heated to 700. K, they react by a second-order process (first order in each reactant), with k = 0.063 L·mol-1s-1 in the rate law (for the rate of formation of HI).
(a) What is the initial reaction rate?
(b) By what factor does the reaction rate increase if the concentration of H2 present in the mixture is tripled?
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