Question

Hi, it is one question with parts (a,b,c,d)

also complete the table

2. A flask is charged with 1.000 mol of A and allowed to react to form B according to the reaction A (g) 2B (g). The following data is obtained for [A] as the reaction proceeds. (10) Complete the table. 0,0 100 200 300 400 Time 0.00 20.0 30.0 40.0 min mol mol 0.835 0.714 0.625 0.555 1.000 [Bl Disappearance of A Appearance of B mol min Avg.rate of 0.0165 mol min Avg. rate of Calculate the rate of disappearance of A between 10.0 and 20.0 s and between 20 and 30 s and 30 and 40s. a. b. What are the [B] at the 5 data points? What are the rates of appearance of B after 10 & 30 min.? c. d. I the reaction rate obeys the rate law, Rate k [A points. ]2, calculate k values at the 10 min and 20 min data

Answer 1

a) Average rate of disappearance of A between 10 and 20 s = (change in concentration of A)/(time elapsed)

= 0.121mol/10s = 0.0121 Mol S^-1

Average rate of disappearance of A between 20 and 30 s = 0.089mol/10s = 0.0089 Mol S^-1

Average rate of disappearance of A between 30 and 40 s = 0.07 mol / 10 s = 0.007 Mol S^-1

b) [B] at each point in time is double the amount of A disappeared.

[B] AT 10 s = 2*0.165 = 0.33 mol

[B] at 20 s = 2*0.286 = 0.572 mol

[B] at 30 s = 2*0.375 = 0.750 mol

[B] at 40 s = 2*0.445=0.890 mol

c) Average rate of appearance of B = 2 times rate of disappearance of A ( According to the stoichiometry)

rate of appearance of B after 10 mins = 2*0.0165=0.033 Mol S^-1

  rate of appearance of B after 30 mins = 2*(0.375/30)=0.025 Mol S^-1

d) average rate of disappearance of A at the 10 s point = 0.0165 Mol S^-1

k = rate/[A]²

= 0.0165/(0.835)² = 0.0236 Mol^-1S^-1

average rate of disappearance of A at the 20 s point = 0.286 mol/20s = 0.0143 Mol S^-1

k = 0.0143/(0.714)² = 0.0280 Mol^-1S^-1