Respuesta :
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.
[tex]\frac{0.088mol-0.110mol}{40s-20s} =-0.0022mol/s[/tex]
B) Consumption.
[tex]\frac{0.054mol-0.088mol}{40s-20s} =-0.0017mol/s[/tex]
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:
[tex]molB_{20s}=0.124mol-0.088mol=0.036molB\\molB_{30s}=0.124mol-0.073mol=0.051molB[/tex]
Now, the rate of appearance:
[tex]r_B=\frac{0.051mol-0.036mol}{30s-20s} =1.5x10^{-3}mol/s[/tex]
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:
[tex]r_B=-r_A=-\frac{0.073mol-0.088mol}{30s-20s} =1.5x10^{-3}mol/s[/tex]
D) Based on the previous analysis:
[tex]molB_{10s}=0.124mol-0.110mol=0.014molB[/tex]
E) Again:
molB_{30s}=0.124mol-0.073mol=0.051molB
Best regards.
The initial number of moles of A of 0.124 moles and the table of values gives the following values;
A) The rate of disappearance of A is 0.0022 moles/s
B) The rate of disappearance of A is 0.0017 moles/s
C) The rate of appearance of B is 0.0015 moles/s
D) 0.014 moles
E) 0.051 moles
How can the average rate of disappearance and number of moles a substance be found from the given table?
A) At 10.0 s, the number of moles present = 0.110 moles
At 20.0 s, the number of moles present = 0.088 moles
Therefore;
[tex]Average \ rate \ of \ disappearance = \mathbf{\dfrac{0.088 - 0.11}{20 - 10}} = -0.0022[/tex]
The average rate of disappearance of A between 10s and 20s is 0.0022 mole/s
B) At 20.0 s, the number of moles present = 0.088 moles
At 40.0 s, the number of moles present = 0.054 moles
Therefore;
[tex]Average \ rate \ of \ disappearance = \mathbf{ \dfrac{0.054 - 0.088}{40 - 20}} = -0.0017[/tex]
The average rate of disappearance of A between 40s and 20s is 0.0017 mole/s
C) At 20.0 s, the number of moles of B present = (0.124 - 0.088) moles = 0.036 moles
At 30.0 s, the number of moles of B present = (0.124 - 0.073) moles = 0.051 moles
Therefore;
[tex]Average \ rate \ of \ disappearance = \dfrac{0.051 - 0.036}{30 - 20} = \mathbf{0.0015}[/tex]
The average rate of appearance of B between 30s and 20s is 0.0015 mole/s
D) The number of moles of B present at 10 s is (0.124 - 0.110) moles = 0.014 moles
E) From the table, the number of moles of B present at 30 s are (0.124 - 0.073) moles = 0.051 moles
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