Hi, it is one question with parts (a,b,c,d) also complete the table

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