Partial pressures and equilibrium?
If initially all partial pressures are .200 atm, what partial pressures are expected at equilibrium?
SO2(g) + NO(2) <---> SO3(g) +NO(g) Kp = 4.0
I don't understand why the pressure changes, is this not already in equilibrium?
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No, its not initially at equilibrium.
Corrected reaction: SO2(g) + NO2(g) <---> SO3(g) +NO(g) Kp = 4.0
Kp = [SO3][NO]/[SO2][NO2]
If you substitute the partial pressures you get:
Qp = (.2)(.2)/(.2)(.2) = 1.0. But the value of Kp is 4.0. This tells you that the system is not at equilibrium.
Equal pressures at the same temperature and volume have the same numbers of molecules.
order is (SO2)(NO2)(SO3)(NO)
Initial: (.2)(.2)(.2)(.2)
change: (-x)(-x)(+x)(+x)
Equilibrium: (0.200 - x)(0.200 -x)(0.200 +x)(0.200 + x)
(0.200 +x)(0.200 + x)/(0.200 - x)(0.200 -x) = 4.0
Now you must solve for x. Then make the mathematical corrections to the concentrations of each reactant and product.
Hope this is helpful to you. JIL HIR