Question

A flask is charged with 0.124 mol of A and allowed to react to from B according to the reaction: A(g) -> B(g). the following drata are obtained for [A] as the reaction proceeds:
time: 0.00 10.0 20.0 30.0 40.0
moles of A: 0.124 0.110 0.088 0.073 0.054
A) what is the average rate of disapperance of A between 10s and 20s?
B) what is the avergae rate of disappearance of A between 20s and 40s?
C) what is the average rate of appearance of B between 20s and 30s?
D) How many moles of B are present at 10s?
E) how many moles of B are present at 30s?

Answer 1

Answer:

A) -0.0022mol/s

B) -0.0017mol/s

C) 1.5x10^{-3}mol/s

D) 0.014molB

E) 0.051molB

Explanation:

Hello, here the answers:

A) Consumption.

$\frac{0.088mol-0.110mol}{40s-20s} =-0.0022mol/s$

B) Consumption.

$\frac{0.054mol-0.088mol}{40s-20s} =-0.0017mol/s$

C) Production.

Since the stoichiometric coefficients are 1 for both A and B, the consumed moles of A are equal to the difference between the initial moles of A and the actual moles of A, thus:

$molB_{20s}=0.124mol-0.088mol=0.036molB\\molB_{30s}=0.124mol-0.073mol=0.051molB$

Now, the rate of appearance:

$r_B=\frac{0.051mol-0.036mol}{30s-20s} =1.5x10^{-3}mol/s$

Which positive indeed because the moles are being formed.

Another way to find is by doing it directly with the moles of A since the rates are numerically equal but differing in the sign:

$r_B=-r_A=-\frac{0.073mol-0.088mol}{30s-20s} =1.5x10^{-3}mol/s$

D) Based on the previous analysis:

$molB_{10s}=0.124mol-0.110mol=0.014molB$

E) Again:

molB_{30s}=0.124mol-0.073mol=0.051molB

Best regards.