3. The water-gas shift reaction is important in several chemical processes, such
ID: 588242 • Letter: 3
Question
3. The water-gas shift reaction is important in several chemical processes, such as the production of H, for fuel cells. This reaction can be written as follows: H,(g) + CO2(g)H20(g) + CO(g) = 0.106 at 700 K. If a mixture of gases that initially contains 0.0150 M Had 0.0150 M CO, is allowed to equilibrate at 700 K, what are the final equilibrium concentrations of all substances present? 4. Atmospheric nitrogen and oxygen react to form nitric oxide N,(g) + O2(g)- 2 NO(g) K, = 2.0 x 10-31 at 25°C. What is the partial pressure of NO in equilibrium with N, and O, atmosphere (P0.78 atm a0.21 atm) in theExplanation / Answer
Answer to problem no(3):
H2(g)+CO2(g)H2O(g)+CO(g)
[H2]
[CO2]
[H2O]
[CO]
initial
0.0150
0.0150
0
0
change
x
x
+x
+x
final
(0.0150 x)
(0.0150 x)
x
x
final
(0.0150 x)
(0.0150 x)
x
x
So, the equilibrium constant for this above reaction
K=[H2O][CO]/[H2][CO2]=(x)(x)/(0.0150x)(0.0150x)=x2/(0.0150x)2=0.106
We could solve this equation with the quadratic formula,
x2/(0.0150x)2=(x/0.0150x)2=0.106
x/(0.0150x)=(0.106)1/2=0.326
x=(0.326)(0.0150)0.326x
1.326x=0.00489
x=0.00369=3.69×103
The final concentrations of all species in the reaction mixture are as follows:
[H2]final =[H2]initial + [H2]=(0.01500.00369) M=0.0113 M
[CO2]final =[CO2]initial + [CO2]=(0.01500.00369) M=0.0113 M
[H2O]final =[H2O]initial + [H2O]=(0+0.00369) M=0.00369 M
[CO] final =[CO]initial + [CO]=(0+0.00369) M=0.00369 M
We can check the equilibrium constant values by inserting the calculated values into the expression:
K=[H2O][CO]/[H2][CO2]=(0.00369)2/(0.0113)2=0.107
H2(g)+CO2(g)H2O(g)+CO(g)
[H2]
[CO2]
[H2O]
[CO]
initial
0.0150
0.0150
0
0
change
x
x
+x
+x
final
(0.0150 x)
(0.0150 x)
x
x
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