At a certain temperature, the equilibrium constant for the following chemical equation is 4.00. At this temperature,...

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Chemistry

At a certain temperature, the equilibrium constant for thefollowing chemical equation is 4.00.
At this temperature, calculate the molarity (M) of NO2(g) thatmust be added to a starting concentration of 2.73 M SO2(g) (with noproducts initially present) in order to have formed 1.30 M SO3(g)and 1.30 M NO(g) when equilibrium is reached.
SO2(g)+NO2(g)âˆ’â‡€â†½âˆ’SO3(g)+NO(g)
molarity =

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4.2 Ratings (713 Votes)

SO2 (g) + NO2 (g) <---------------> SO3 (g) + NO (g)

2.73 Â Â Â Â Â Â Â Â Â Â Â Â x Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â 0 ----------------------> initial

-yÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â - yÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +yÂ Â Â Â Â Â Â Â Â Â +y --------------------> dissociation

2.73 - y Â Â Â Â Â x - y Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â y Â Â Â Â Â Â Â Â Â y-----------------------> equilibrium

2.73-1.3Â Â Â Â Â x-1.3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1.3Â Â Â Â Â Â Â Â Â 1.3Â Â Â

1.43 Â Â Â Â Â Â Â Â Â x-1.3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1.3Â Â Â Â Â Â Â Â Â 1.3 (afetr below calculation)

at equilibrium SO3 moles given so

y = 1.3

Kc = 4

Kc = [ SO3][NO]/[SO2][NO2]

4= (1.3 )^2 / (1.43) (x - 1.3)

x = 1.6

NO2 moles = x = 1.6

molarity of NO2 = 1.60 M

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