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

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation: