K is 0.87 at a certain temperature. A 5.0-L flask at equilibrium is determined t
ID: 980998 • Letter: K
Question
K is 0.87 at a certain temperature. A 5.0-L flask at equilibrium is determined to have a total pressure of 1.25 atm an oxygen to have a partial pressure of 0.515 atm. Calculate Pno and PNO_2 at equilibrium. Consider the following reaction: At a certain temperature, the equilibrium constant for the reaction is 0.0639. What are the partial pressures of all gases at equilibrium if the initial partial pressure of the gases (both products and reactants) is 0.400 atm? Consider the hypothetical reaction at 325degreeC What are the equilibrium partial pressures of all the gases if all the gases (products and reactants) have an initial partial pressure of 0.228 atm?Explanation / Answer
ANSWER
N2 (g) + O2 (g) <-------> 2NO (g)
I 0.4 0.4 0.4
C - x - x +x
E 0.4-x 0.4-x 0.4+x
0.0639 = (0.4 + x)2 / (0.4-x)(0.4-x)
0.0639 = (0.4 + x)2 / (0.4-x)2
Upon solving the above equation it takes the for of a quadraitic equation
0.9361x2 + 0.85x + 0.15 = 0
x = -0.22
Therefore partial pressure of the gases at equilibrium is given below
PN2 = 0.4 - (-0.22) = 0.62atm
PO2 = 0.4 - (-0.22) = 0.62atm
PNO = 0.4 + (-0.22) = 0.18atm
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