assuming oxygen behaves like an ideal gas, what volume in liters would 3.50 moles of oxygen gas occupy at STP?
1 answer:
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Answer:
a. Kp=1.4
b.Kp=2.0 * 10^-4
c.Kp=2.0 * 10^5
Explanation:
For the reaction
A(g)⇌2B(g)
Kp is defined as:
The conditions in the system are:
A B
initial 0 1 atm
equilibrium x 1atm-2x
At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.
Replacing these values in the expression for Kp we get:
Working with this equation:
This last expression is quadratic expression with a=4, b=-(4+Kp) and c=1
The general expression to solve these kinds of equations is:
(equation 1)
We just take the positive values from the solution since negative partial pressures don´t make physical sense.
Kp = 1.4
With x1 we get a partial pressure of:
Since negative partial pressure don´t make physical sense x1 is not the solution for the system.
With x2 we get:
These partial pressures make sense so x2 is the solution for the equation.
We follow the same analysis for the other values of Kp.
Kp=2*10^-4
X1=0.505
X2=0.495
With x1
Not sense.
With x2
X2 is the solution for this equation.
Kp=2*10^5
X1=50001
With x1
Not sense.
With x2
X2 is the solution for this equation.