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kakasveta [241]
3 years ago
8

How does the value of kc in n2o4(g)⇋2no2(g) kc=[no2]2[n2o4] depend on the starting concentrations of no2 and n2o4?

Chemistry
1 answer:
Vlad [161]3 years ago
8 0
<h3><u>Answer;</u></h3>

The value of Kc does not depend on starting concentrations.

<h3><u>Explanation;</u></h3>
  • <em><u>At constant temperature, changing the equilibrium concentration does not affect the equilibrium constant, because the rate constants are not affected by the concentration changes. </u></em>
  • When the concentration of one of the participants is changed, the concentration of the others vary in such a way as to maintain a constant value for the equilibrium constant.
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Supposing a temperature of 25 degrees and supposing that all activity coefficients are 1 

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Which of the following is true regarding the law of conservation of mass
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The initial pressure of a mixture of C6H6 and an excess of H2 in a rigid vessel is 1.21 atm. A catalyst is introduced. After the
KengaRu [80]

Answer:

mole fraction of C6H6 = 0.613 atm

Explanation:

The equation for this reaction is :

          C_6H_6 _{(g)} + 3H_2_{(g)} \to C_6H_{12}_{(g)}

Initial      P₁            P₂             0

Final        0           P₂ -P₁/2      P₁

After completion of the reaction;

P₁  +  P₂  = 1.21 atm                    ----- (1)

P₂  -  P₁/2 +  P₁ = 0.839 atm

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Subtracting (2) from (1); we have:

P₁/2 = 0.371

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From(1)

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0.742 atm + P₂  = 1.21 atm

P₂  = 1.21 atm - 0.742 atm

P₂  = 0.468 atm

Thus, the partial pressure of C6H6 = 0.742 atm

∴

Partial pressure = Total pressure × mole fraction of C6H6

mole fraction of C6H6 = Partial pressure /  Total pressure

mole fraction of C6H6 = 0.742 atm / 1.21 atm

mole fraction of C6H6 = 0.613 atm

5 0
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