Question

The equilibrium constant for the reaction 2NO(g)+Br2(g)⥫⥬==2NOBr(g) is Kc=1.3×10−2 at 1000 K. At this temperature does the equilibrium favor NO and Br2, or does it favor NOBr? Calculate Kc for 2NOBr(g)⥫⥬==2NO(g)+Br2(g). Calculate Kc for NOBr(g)⥫⥬==NO(g)+12Br2(g).

Answer 1

Answer :

The equilibrium favors NO and $Br_2$.

(1) The value of equilibrium constant for this reaction is, 76.9

(2) The value of equilibrium constant for this reaction is, 8.77

Explanation:

The given chemical equation is:

$2NO(g)+Br_2(g)\rightarrow 2NOBr(g)$

The value of equilibrium constant for the above equation is $K_c=1.3\times 10^{-2}$.

The value of $K_c$ that means equilibrium lies to the left side. Thus, the equilibrium favors NO and $Br_2$.

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

(1) $2NOBr(g)\rightarrow 2NO(g)+Br_2(g)$

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

The value of equilibrium constant for reverse reaction is:

$K_{c_1}=\frac{1}{K_c}$

$K_{c_1}=\frac{1}{1.3\times 10^{-2}}=76.9$

Thus, the value of equilibrium constant for this reaction is, 76.9

(2) $NOBr(g)\rightarrow NO(g)+\frac{1}{2}Br_2(g)$

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

If the equation is multiplied by a factor of '1/2', the equilibrium constant of the reverse reaction will be the 1/2 power of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{c_2}=(\frac{1}{K_c})^{1/2}$

$K_{c_2}=(\frac{1}{1.3\times 10^{-2}})^{1/2}=8.77$

Thus, the value of equilibrium constant for this reaction is, 8.77

Answer 2

Answer:

See explanation below

Explanation:

In order to do this, let's write again the equation:

2NO(g)+Br₂(g)⥫⥬==2NOBr(g)   Kc = 1.3x10⁻²

Now, we want to know if this reaction favors reactants or product. Let's see the value and expression of Kc.

Kc = [NOBr]² / [NO]² [Br₂] = 1.3x10^-2

If the value of Kc is <1, means that the result of the multiplication of the products in the denominator, is a number that is greater than the number given in the numerator, therefore, a number that is smaller than the denominator will result in a number <1, so, in other words, the reactants are interacting even more and are exerting a greater concentration to favor the formation of these reactants in equilibrium. That's why the reaction at this temperature and conditions, will favor the reactants NO and Br₂

Now, we want to know the new value of Kc for the reverse reaction which is:

2NOBr(g) <---------> 2NO(g) + Br₂(g)       Kc2 = ?

We know by the previous reaction that the expression for Kc (From reaction 1) is:

Kc1 = [NOBr]² / [NO]² [Br₂]   (1)

The Kc2 for the above reaction is:

Kc2 = [NO]² [Br₂] / [NOBr]²

So, in order to know the value for Kc2, all we need to do is convert the expression of Kc1 into Kc2. How can we do this? matemathically speaking, the only way to inverse the value of an operation is doing a division with 1 in this way:

1/Kc1 = 1 / [NOBr]² / [NO]² [Br₂]

1/Kc1 = [NO]² [Br₂] / [NOBr]²   and this is Kc2 so:

Kc2 = 1/Kc1

All we have to do is replace the value and solve for Kc2:

Kc2 = 1 / 1.3x10⁻²

Kc2 = 76.92

Finally, for the last reaction which is:

NOBr(g)⥫⥬==NO(g)+1/2Br₂(g).   Kc3 = ?

This reaction has the difference with the previous one, in the fact that it's coefficients that balance the equation are reduce to half, so, in order to calculate Kc3, we do mathematical operations. Let's write Kc2 and Kc3, and then, let's obtain Kc3 with the operations:

Kc2 = [NO]² [Br₂] / [NOBr]²

Kc3 = [NO] [Br₂]¹/² / [NOBr]

So, how can we convert Kc2 into Kc3?, simply, we just do root square to the expression of Kc2 so:

√Kc2 = √[NO]² √[Br₂] / √[NOBr]²

With these, the root cancels the power so:

√Kc2 = [NO] √[Br₂] / [NOBr]

In other words, we already convert Kc2 in Kc3 and in conclusion we have:

Kc3 = √Kc2

Replace the values:

Kc3 = √76.92

Kc3 = 8.77