ashkaputrilma
contestada

The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant.
2 HD(g) ⇌ H2(g) + D2(g)
Kc = 0.28

6 H2(g) + 6 D2(g) ⇌ 12 HD(g)
Kc = ?

A) 0.00048
B) 1.2
C) 1.62
D) 2075
E) 0.81

Respuesta :

meerkat18
given the original equation of 2 HD(g) ⇌ H2(g) + D2(g)  with Kc = 0.28 and asked for the kc of the equation 6 H2(g) + 6 D2(g) ⇌ 12 HD(g), we can see that the equation 1 is reversed and multiplide by 6. in this case, the reverse means kc is equal to 1/0.28 and that the multiplication by 6 raises the new kc to 6. the new kc is equal to D. 2075
Blitztiger
The equilibrium constant is equal to concentration products raised to their stoichiometric coefficients, divided by the same for the reactants.

For the first reaction: [tex]K_c=\frac{[H_2][D_2]}{[HD]^2}=0.28[/tex]

For the second reaction: [tex]K_c=\frac{[HD]^{12}}{[H_2]^6[D_2]^6}=(\frac{[H_2][D_2]}{[HD]^2})^{-6}=(0.28)^{-6}=2075[/tex]
This is choice D.

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