Lconsider the following reaction: ch3br(aq)+oh−(aq)−→−ch3oh(aq)+br−(a q) the rate law for this reaction is first order in ch3br and first order in oh−. when [ch3br] is 5.0×10−3m and [oh−] is 0.050 m, the reaction rate at 298 k is 0.0432m/s. (a) what is the value of the rate constant? (b) what are the units of the rate constant? (c) what would happen to the rate if the concentration of oh− were tripled? (d) what would happen to the rate if the concentration of both reactants were tripled?

we need the questions , i may then you.

the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole