Question

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

Answer 1

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

Answer 2

The equilibrium constant is equal to concentration products raised to their stoichiometric coefficients, divided by the same for the reactants.

For the first reaction:
`K_c=([H_2][D_2])/([HD]^2)=0.28`

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