A flask is charged with 0.130 mol of A and allowed to react to form B according
ID: 475877 • Letter: A
Question
A flask is charged with 0.130 mol of A and allowed to react to form B according to the following hypothetical gas-phase reaction.
The following data are collected.
(a) Calculate the number of moles of B at each time in the table.
0 s
mol
40 s
mol
80 s
mol
120 s
mol
160 s
mol
(b) Calculate the average rate of disappearance of A for each 40 s interval, in units of mol/s.
0 - 40 s
mol/s
40 - 80 s
mol/s
80 - 120 s
mol/s
120 - 160 mol/s
mol/s
Explanation / Answer
(a) Calculate the number of moles of B at each time in the table.
A(g) B(g)
0 s
mol
At 0 secone there is no B thus the mole so f B is 0.
40 s
mol
A(g) B(g)
I 0.130 0
At 40s 0.071 x
X= 0.130 -0.071= 0.059 mole
80 s
mol
A(g) B(g)
I 0.130 0
At 40s 0.044 x
X= 0.130 -0.044 = 0.086 mole
120 s
mol
A(g) B(g)
I 0.130 0
At 40s 0.029 x
X= 0.130 -0.029= 0.101 mole
160 s
mol
A(g) B(g)
I 0.130 0
At 40s 0.021 x
X= 0.130 -0.021= 0.109 mole
times (s)
0
40
80
120
160
moles of A
0.130
0.071
0.044
0.029
0.021
(b) Calculate the average rate of disappearance of A for each 40 s interval, in units of mol/s.
0 - 40 s
mol/s
the average rate of disappearance of A = A final - initial
= 0.071-0.130
= -0.059
= -0.059 mole /40 sec
= -0.001475 mole s-1
40 - 80 s
mol/s
the average rate of disappearance of A = A final - initial
= 0.044-0.071
= - 0.027
= -0.027 mole /40 sec
= -0.000675 mole s-1
80 - 120 s
mol/s
the average rate of disappearance of A = A final - initial
= 0.029- 0.044
= - 0.015
= -0.015 mole /40 sec
= -0.000375 mole s-1
120 - 160 mol/s
mol/s
the average rate of disappearance of A = A final - initial
= 0.021- 0.029
= - 0.008
= -0.008 mole /40 sec
= -0.0004 mole s-1
times (s)
0
40
80
120
160
moles of A
0.130
0.071
0.044
0.029
0.021
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