Respuesta :
Answer: [tex]K_c[/tex] for this reaction at this temperature is 0.029
Explanation:
Moles of [tex]HBr[/tex] = 2.00 mole
Volume of solution = 4.00 L
Initial concentration of [tex]HBr=\frac{moles}{Volume}=\frac{2.00}{4.00L}=0.500M[/tex]
The given balanced equilibrium reaction is,
[tex]2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)[/tex]
Initial conc. 0.500 M 0 M 0 M
At eqm. conc. (0.500-2x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[H_2\times [Br_2]}{[HBr]^2}[/tex]
Equilibrium concentration of [tex][Br_2][/tex] = x = 0.0955 M
Now put all the given values in this expression, we get :
[tex]K_c=\frac{0.0955\times 0.0955}{0.500-2\times 0.0955}[/tex]
[tex]K_c=0.029[/tex]
Thus [tex]K_c[/tex] for this reaction at this temperature is 0.029
The Kc for this reaction at this temperature is 0.0955.
Calculation of the Kc:
Since
Initial moles of HBr,n = 2.00 mol
The volume of the container, V = 4.00 L
So,
initial concentration of HBr
= n/V = 2.00mol/4.00L
= 0.50 M
And, The concentration of Br2 at equilibrium = 0.0955 M
The given chemical reaction at equilibrium is
2HBr(g) <=============> H2(g) + Br2(g)
Initial conc. 0.50M 0M 0M
Conc. at equilibrium: (0.50 - 2x0.0955)M 0.0955M 0.0955M
= 0.309M
So,
Hence, [H2(g)] = 0.0955M, [Br2(g)] = 0.0955M
Now
[2HBr(g)] = 0.309 M
And, finally Kc can be calculated as
Kc = [H2(g)] x [Br2(g)] / [2HBr(g)]2
= 0.0955M x 0.0955M / (0.309M)2
= 0.0955
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