The equilibrium constant is given for one of the reactions below. determine the value of the missing equilibrium constant. 2 hd(g)

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

Answers

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=frac{[H_2][D_2]}{[HD]^2}=0.28

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

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

- Kc = 6.63

Solution:- First given equation is:

N_2O_4(g)rightleftharpoons 2NO_2(g)

The equilibrium expression for this equation is written as:

Kc=frac{[NO_2]^2}{[N_2O_4]}

The second given equation is:

5N_2O_4(g)rightleftharpoons 10NO_2(g)

The equilibrium expression for this equation is written as:

Kc^'=frac{[NO_2]^1^0}{[N_2O_4]^5}

Comparing these two expressions, to get this second expression from the first one, we need to do the fifth power of first equilibrium expression.

Hence, Kc^'=(Kc)^5

Plug in the value of Kc in it from first expression,  Kc^'=(1.46)^5

Kc^'  = 6.63

So, the Kc of second equation is 6.63.

12.7551

Explanation:

The given chemical equation follows:

2HD(g)rightarrow H_2(g)+D_2(g)

The equilibrium constant for the above equation is 0.28.

We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:

2H_2(g)+2D_2(g)rightarrow 4HD(g)

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

If the equation is multiplied by a factor of '2', the equilibrium constant of the reverse reaction will be the square of the equilibrium constant  of initial reaction.

The value of equilibrium constant for reverse reaction is:

K_{eq}'=(frac{1}{0.28})^2

Hence, the value of equilibrium constant for reverse reaction is 12.7551.

Answer choice 'D' => Kc(2) = 581

Explanation:

2HD ⇄ H₂ + D₂ => Kc(1) = [H₂][D₂]/[HD]² => 0.28

5H₂ + 5D₂ ⇄ 10HD => Kc(2) = [HD]¹⁰/[H₂]⁵[D₂]⁵ = 1/(Kc(1))⁵ = 1/(0.28)⁵ = 581

For these type problems, one should 1st write the empirical Kc expression for each equation given. Then compare the expressions and ask 'how can the 1st Kc expression be changed into the 2nd Kc expression?' Apply and substitute given Kc(1) into the Kc(2) expression and solve for numerical results.  

Kc is simply the ratio of the concentration of the product over the reactant, that is:

Kc = [NO2]^2 / [N2O4] = 1.46

 

Now since the overall equation or concentration is multiplied by 5, therefore:

{[NO2]^2 / [N2O4]}^5 = 1.46^5 =>

Kc = 6.63

The answer is C) 6.63

Explanation:

When the coefficients in a balanced equation are multiplied by a factor, the resulting equilibrium constant is raised to this factor.

In this case, you have the following equilibrium:

N₂O₄(g) ⇌ 2 NO₂(g) Kc = 1.46

In which the coefficients are 1 (for N₂O₄) and 2 (for NO₂)

In the second equation, notice that you have the coefficients of the first equation multiplied by 5 (1 x 5= 5 for N₂O₄; 2 x 5= 10 for NO₂) :

5 N₂O₄(g) ⇌ 10 NO₂(g)

Thus, to obtain the equilibrium constant of the second equation (Kc'), you have to raise the first equilibrium constant (Kc) to 5:

K'c= (Kc)⁵= (1.46)⁵= 6.63

To calculate Kc of the second step, we must observe the first reaction.
If we rotate the first reaction and multiplay it by 2 we get the same thing.

So the rotation means 1/kc(first reaction)
And by multiplying it by 2, K should be squared.

Kc2=1/(kc1)^2=6.92*10^-10

 the value of the missing equilibrium constant is E. 7.30
Solution:
N2O4(g) ⇌ 2 NO2(g)Kc = 1.46
5 N2O4(g) ⇌ 10 NO2(g)Kc =?

If 1 atom of N2O4(g) is 1.46
So to get 5 atoms of N2O4(g), you will simply multiply it by 5.

1.46 x 5 = 7.30

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