Could someone help me with this lab advance study! Show work please? A student s
ID: 969943 • Letter: C
Question
Could someone help me with this lab advance study! Show work please?
A student studied the clock reaction described in this experiment. She set up Reaction Mixture 2 by mixing 20 mL 0.010 M KI, 10 mL 0.001 M Na_2S_2,O_3. 10 0.040 M KBrO_3 and 10 mL 0.10 M HCI using the procedure given. It took about 45 seconds for the color to turn blue. a. She found the concentrations of each reactant in the reacting mixture by realizing that the number of moles of each reactant did not change when that reactant was mixed with the others, but that its concentration did. For any reactant A, no. moles A = M_A stock times v_stock = M_A mixture x V_mixture The volume of the mixture was 50 mL. Revising the above equation, she obtained M_A mixture = M_A stock times V_stock(mL)/50mL Find the concentrations of each reactant by using the above equation. (I^-) = M; (BrO_3^-) = M; (H^+) = M b. What was the relative rate of the reaction (1000/t)? c. Knowing the relative rate of reaction for Mixture 2 and the concentrations of I^-, BrO_3^-. and H^+ in that mixture, she was able to set up Equati On 5 for the relative rate of the reaction. The only quantities that remained unknown were if, m, n, and p. Set up Equation 5 as she did, presuming she did it properly. Equation 5 is on page 169. 2. For Reaction Mixture 1 the student found that 85 seconds were required. On dividing Equation 5 for Reaction Mixture 1 by Equation 5 for Reaction Mixture 2, and after canceling out the common terms (k', terms in (Bro_3^-) and (H^+)), she got the following equation: 11.8/22 = (0.0020/0.0040)^n = (1/2)^n Recognizing that 11.8/22 is about equal to 1/2, she obtained an approximate value for m. What was that value? m = By taking logarithms of both sides of the equation, she got an exact value for m. What was that value? m = Since orders of reactions are often integers, she reported her approximate value as the order of the reaction with respect to I^-. 137Explanation / Answer
Concentration of KI= 0.010*20/50= 0.004 M (I-)
Concentration of Na2SO4= 0.001*10/50=0.0002M
Concentration of KBrO3= .040*10/50= 0.008M (BrO3-)
Concentration of HCl = 0.1*10/50= 0.02M (H+)
Relative rate= 0.004/45 =8.88*10-5 M/Sec
The order m=1