The partial pressure of reaction is 1.017*10-3 at 500c. Calculate kp at 600c
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Answer:- 0.025 atm
Solution:- The given balanced equilibrium equation is:
2NO_2(g)\leftrightarrow 2NO(g)+O_2(g)2NO2(g)↔2NO(g)+O2(g)
From the equation, there is 2:1 mole ratio between NO and O_2O2 .
Their equilibrium partial pressure would be in the ratio. Equilibrium partial pressure of oxygen is given as 0.25 atm. So, the equilibrium partial pressure of NO = 2(0.25 atm) = 0.50 atm
Equilibrium expression for the given equation is written as:

Let's plug in the values in it and solve it for equilibrium partial pressure of NO_2NO2 . Let's say the equilibrium partial pressure of nitrogen dioxide is p.
100=\frac{(0.50)^2(0.25))}{p^2}100=p2(0.50)2(0.25))
100=\frac{0.0625}{p^2}100=p20.0625
On rearranging the above equation:
p^2=\frac{0.0625}{100}p2=1000.0625
On taking square root to both sides:
p=\frac{0.25}{10}p=100.25
p = 0.025 atm
So, the equilibrium partial pressure of NO_2NO2 is 0.025 atm.
Solution:- The given balanced equilibrium equation is:
2NO_2(g)\leftrightarrow 2NO(g)+O_2(g)2NO2(g)↔2NO(g)+O2(g)
From the equation, there is 2:1 mole ratio between NO and O_2O2 .
Their equilibrium partial pressure would be in the ratio. Equilibrium partial pressure of oxygen is given as 0.25 atm. So, the equilibrium partial pressure of NO = 2(0.25 atm) = 0.50 atm
Equilibrium expression for the given equation is written as:

Let's plug in the values in it and solve it for equilibrium partial pressure of NO_2NO2 . Let's say the equilibrium partial pressure of nitrogen dioxide is p.
100=\frac{(0.50)^2(0.25))}{p^2}100=p2(0.50)2(0.25))
100=\frac{0.0625}{p^2}100=p20.0625
On rearranging the above equation:
p^2=\frac{0.0625}{100}p2=1000.0625
On taking square root to both sides:
p=\frac{0.25}{10}p=100.25
p = 0.025 atm
So, the equilibrium partial pressure of NO_2NO2 is 0.025 atm.
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