A flask is charged with 0.124 mol of A and allowed to react to form B according to the reaction A(g) →B(g). The following data are
obtained for [ A ] as the reaction proceeds:
Time (s) 1 10 20 30 40
Moles of A 0.124 0.110 0.088 0.073 0.054
how do i solve for B?
for problem 8: also i know how to solve for A but i cant figure out the one below, i tried subtracting
0.088 - 0.073/20 = but its wrong
7) The average rate of appearance of B between 20 s and 30 s is __________ mol/s.
A) +7.3 × 10-3
B) +1.5 × 10-3
C) -7.3 × 10-3
D) +5.0 × 10-4
E) -1.5 × 10-3
7)
8) The average rate disappearance of A between 20 s and 30 s is __________ mol/s.
A) 1.6 × 10-2
B) 0.15
C) 670
D) 1.5 × 10-3
E) 5.0 × 10-4
To solve for B, you need to calculate the change in the moles of A and use the stoichiometry of the reaction A(g) → B(g) to determine the number of moles of B produced.
Here, the change in moles of A is given by:
Δ[A] = [A]final - [A]initial
Δ[A] = 0.054 - 0.088
Δ[A] = -0.034 mol
According to the balanced equation, the stoichiometric ratio between A and B is 1:1. Therefore, the number of moles of B produced is also -0.034 mol.
Now, let's solve problem 7:
The average rate of appearance of B between 20 s and 30 s can be calculated using the formula:
Average rate = Δ[B]/Δt
Here, Δ[B] = -0.034 mol (since the number of moles of B produced is -0.034 mol), and Δt = 30 s - 20 s = 10 s.
Average rate = -0.034 mol / 10 s
Average rate = -0.0034 mol/s
Since the question asks for the answer in mol/s (rate of appearance is positive), the correct option is:
A) +7.3 × 10-3
Now, let's solve problem 8:
The average rate disappearance of A between 20 s and 30 s can be calculated using the formula:
Average rate = Δ[A]/Δt
Here, Δ[A] = -0.088 mol (as given in the data), and Δt = 30 s - 20 s = 10 s.
Average rate = -0.088 mol / 10 s
Average rate = -0.0088 mol/s
Since the question asks for the answer in mol/s, the correct option is not provided in the given options.
Please double-check the options provided for problem 8 as they seem to be incorrect or missing the correct answer.