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3 years ago

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A 1.00 L flask was filled with 2.00 mol gaseous SO2 and 2.00 mol gaseous NO2 and heated. After equilibrium was reached, it was f

ound that 1.30 mol gaseous NO was present. Assume that the reaction occurs under these conditions. Calculate the value of the equilibrium constant, K, for the following reaction. SO2(g) NO2(g) equilibrium reaction arrow SO3(g) NO(g)

1 answer:

Nataly_w [17]3 years ago

6 0

Answer:

Explanation:

SO3 (g) + NO (g) U SO2 (g) + NO2 (g)

occurs under these conditions. Calculate the value of the equilibrium constant, Kc, for the above reaction.

SO3 (g) + NO (g) U SO2 (g) + NO2 (g)

Initial (M) 2.00 2.00 0 0

Change (M) −x −x +x +x

Equil (M) 2.00 − x 2.00 − x x x

2 2 c 3

2

c 2

[SO ][NO ]

[SO ][NO]

(2.00 )

=

= −

K

x K

x

Since the problem asks you to solve for Kc, it must indicate in the problem what the value of x is. The concentration of

NO at equilibrium is given to be 1.30 M. In the table above, we have the concentration of NO set equal to 2.00 − x.

2.00 − x = 1.30

x = 0.70

Substituting back into the equilibrium constant expression:

2c 2 2c 2

(2.00 )

(0.70)

(2.00 0.70)

= − = −

x KxK

Kc = 0.290

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${\rm ?\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}$ .

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By the conservation of atoms, the number of atoms on the right-hand side of the reaction should match those on the left-hand side. In this reaction, $\rm CO_2$ is the only product with carbon atoms, whereas $\rm H_2O$ is the only product with hydrogen atoms. These $3$ carbon atoms and $8$ hydrogen atoms would correspond to:

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$Molarity=\frac{1.45 moles}{2.59 L}$

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<u><em>The molarity is 0.56</em></u> $\frac{moles}{L}$ <u><em></em></u>

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