Asked by arshly
The equilibrium constant, Keq, for the reaction:
SO3(g) + NO (g) ⇋ NO2 (g) + SO2 (g)
was found to be 0.500 at a certain temperature. If 0.300 mol of SO3 and 0.300 mol NO were placed in a 2.00 L container and allowed to react, what would be the equilibrium concentration of each gas?
Answers
Answered by
DrBob222
(SO2) = 0.300 mol/2.00 L = 0.150 = (NO)
....................SO3(g) + NO (g) ⇋ NO2 (g) + SO2 (g)
I..................0.150 ......0.150............0..............0
C...................-x..............-x...............x..............x
E...............0.150-x.....0.150-x...........x...............x
Keq = 0.500 = (NO2)(SO2)/(SO3)(NO)
Substitute the E line into the Keq expression and solve for x, then evaluate each material. Post your work if you get stuck.
....................SO3(g) + NO (g) ⇋ NO2 (g) + SO2 (g)
I..................0.150 ......0.150............0..............0
C...................-x..............-x...............x..............x
E...............0.150-x.....0.150-x...........x...............x
Keq = 0.500 = (NO2)(SO2)/(SO3)(NO)
Substitute the E line into the Keq expression and solve for x, then evaluate each material. Post your work if you get stuck.
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