Calculate the value of the equilibrium constant, K c , for the reaction Q ( g ) + X ( g ) − ⇀ ↽ − 2 M ( g ) + N ( g ) given that M ( g ) − ⇀ ↽ − Z ( g ) K c 1 = 3.15 6 R ( g ) − ⇀ ↽ − 2 N ( g ) + 4 Z ( g ) K c 2 = 0.509 3 X ( g ) + 3 Q ( g ) − ⇀ ↽ − 9 R ( g ) K c 3 = 12.5

The value of the equilibrium constant is 0.167

Explanation:

Step 1: The target equation

Q(g) + X(g) ⇔ 2M(g) + N(g)

Given is:

(1) M(g)⇔Z(g)   c1=3.15

(2) 6R(g) ⇔ 2N(g) + 4Z(g)   c2=0.509

(3) 3X(g) +3Q(g) ⇔ 9R(g)    c3=12.5

Step 2: Rearange the equation

We have to rearange the equation to come to the final result

This is Hess' Law

In the target equation we have Q(g) + X(g)

In (3)  we have 3X(g) +3Q(g) ⇔ 9R(g)

To get the target of  Q(g) + X(g) we have to divide (3) by 3. This will give us:

X(g) +Q(g) ⇔ 3R(g)   Kc = ∛12.5 = 2.32  (Note: to get Kc of the target equation we use cube root)

The target equation has as product  2M(g) + N(g)

To get M(g) we will use the (1) equation

Since M(g) is a product and not a reactant, we have to reverse the equation. Next to that we also have to double the equation because we need 2M(g) and not M(g)

2Z(g) ⇔ 2M(g)        Kc = 1/(3.15)²  = 0.101   (Note: to get Kc' after reversing the equation we calculate 1/Kc.   To get Kc'' after doubling and reversing the equation we calculate 1/(Kc²)

To get N(g) we will use (2) 6R(g) ⇔ 2N(g) + 4Z(g)

Since we only need N(g) we will divide this equation by 2. This will get us:

3R(g) ⇔ N(g) + 2Z(g)    Kc = √0.509   = 0.713   (Note: if we divide the equation by 2, to calculate Kc' we use square root)

Now we have all the components we will add the 3 equations:

X(g) +Q(g) + 2Z(g)  + 3R(g)⇔ 3R(g) + 2M(g) + N(g) + 2Z(g)

We will simplify this equation:

X(g) +Q(g) ⇔  2M(g) + N(g)  this is our target equation

The value of the  equilibrium constant, Kc is:

Kc = 2.32 * 0.101*0.713

Kc = 0.167

Note: to calculate Kc after adding several equations,we'll multiply Kc1* Kc2 * Kc3 etc...

The value of the equilibrium constant is 0.167  

Explanation:

We have three equations:

1. M(g) ⇌ Z(g);                      Kc₁ =    3.15

2. 6R(g) ⇌ 2N(g) + 4Z(g); Kc₂ =    0.509

3. 3X(g) + 3Q(g) ⇌ 9R(g); Kc₃ = 12.5

From these, we must devise the target equation:

4. Q(g) + X(g) ⇌ 2M(g) + N(g); Kc = ?

The target equation has Q(g) on the left, so you divide Equation 1 by 3.

When you divide an equation by 3, you take the cube root of its Kc.

5. X(g) + Q(g) ⇌ 3R(g): K₅ = ∛(Kc₃)

Equation 5 has 3R on the right, and that is not in the target equation.

You need an equation with 3R on the left, so you divide Equation 2 by 2.  

When you divide an equation by 2, you take the square root of its Kc.

6. 3R(g) ⇌ N(g) + 2Z(g); K₆ = √ (Kc₂)

Equation 6 has 2Z on the right, and that is not in the target equation.

You need an equation with 2Z on the left, so you reverse Equation 2 by and double it.

When you reverse an equation, you take the reciprocal of its K.

When you double an equation, you square its K.

7. 2Z(g) ⇌ 2M(g); K₇ = (1/Kc₁)²

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you multiply their K values.

You get the target equation 4:

5. X(g) + Q(g) ⇌ 3R(g);              K₅ = ∛(Kc₃)

6. 3R(g) ⇌ N(g) + 2Z(g);             K₆ = √(Kc₂)

7. 2Z(g) ⇌ 2M(g);                        K₇ = (1/Kc₁)²

4. Q(g) + X(g) ⇌ 2M(g) + N(g); Kc = K₅K₆K₇ =  [∛(Kc₃)√(Kc₂)]/(Kc₁)²

Kc =  [∛(12.5)√(0.509)]/(12.5)² = (2.321 × 0.7120)/156.2 = 0.0106

b is the correct answer

mark me brainliest and hope this