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Bezzdna [24]
2 years ago
10

Consider the reaction shown.

Chemistry
1 answer:
Allushta [10]2 years ago
6 0

Answer:

4.43 g Cl₂

Explanation:

To find the mass of Cl₂, you need to (1) convert moles HCl to moles Cl₂ (via the mole-to-mole ratio from equation coefficients) and then (2) convert moles Cl₂ to grams (via the molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units. The final answer should have 3 significant figures like the given value.

4 HCl(g) + O₂(g) -----> 2 Cl₂(g) + 2 H₂O(g)
^                                   ^

Molar Mass (Cl₂): 2(35.453 g/mol)

Molar Mass (Cl₂): 70.906 g/mol

0.125 moles HCl          2 moles Cl₂             70.906 g
--------------------------  x  ----------------------  x  -------------------  =  4.43 g Cl₂
                                     4 moles HCl              1 mole

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Answer : The equilibrium constant K_c for the reaction is, 0.1133

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\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}

\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M

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The balanced equilibrium reaction is,

                              Br_2(g)\rightleftharpoons 2Br(aq)

Initial conc.         1.731 M      0

At eqm. conc.      (1.731-x)    (2x) M

As we are given,

The percent of dissociation of Br_2 = \alpha = 1.2 %

So, the dissociate concentration of Br_2 = C\alpha=1.731M\times \frac{1.2}{100}=0.2077M

The value of x = 0.2077 M

Now we have to calculate the concentration of Br_2\text{ and }Br at equilibrium.

Concentration of Br_2 = 1.731 - x  = 1.731 - 0.2077 = 1.5233 M

Concentration of Br = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

K_c=\frac{[Br]^2}{[Br_2]}

Now put all the values in this expression, we get :

K_c=\frac{(0.4154)^2}{1.5233}=0.1133

Therefore, the equilibrium constant K_c for the reaction is, 0.1133

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