# 2-4 please (Sorry, I\'m not sure how to enlarge the picture ... I tried on my

Question

# 2-4 please (Sorry, I'm not sure how to enlarge the picture ... I tried on my phone and laptop)

The following rate data was determined for the following ran: S_2O_8^2 + 3I rightarrow 2SO_3^2- + I_3^- Find the rate law for the ran: determine k for the ran (with correct units); and find the rate of disappearance of I- when [S_2O_8^2] = 0.075M and [I^-] = 0.050M. The rate constant for a 1^st order decomposition of an insecticide in water is 1.45yr^-1. This insecticide is spilled into a lake on June 1^st at a concentration of 5.0 times 10^-2 g/mL. How long will it take for the concentration to drop to 3.0 times 10^-7 g/mL? What's the half-life of the insecticide? The following ran is 2^nd order with respect to NO_3. The initial [NO_2] = 0.400M and its concentration drops to 0.220 M at t = 10 minutes. 2 NO_2 rightarrow 2 NO + O_2 Find k; and find [NO_2] after 20 minutes.

Explanation / Answer

2)    Given that k = 1.45yr1

For first order recation,

k = 1/t ln { [A]o/[A]t} -----Eq (2)

Given that

Initial concentration [A]o = 5 x 10-3 g/mL

Final concentration [A]t = 3 x 10-7 g/mL

time t = ?

k = (1/t) ln {[A]o/ [A]t}

1.45 = (1/t ) ln {5 x 10-3/ 3 x 10-7}

t = ln {5 x 10-3/ 3 x 10-7} / 1.45

   = 6.7 yrs

t = 6.7 yrs

Therefore, it will take 6.7 yrs to drop the concentration to 3 x 10-7 g/mL.

3) For 1st order decomposition,

            half life= 0.693/ rate constant

Therefore,

     Half life of insecticide = 0.693 / k

                                    = 0.693 / 1.45 yr-1

                                    = 0.48 yrs

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