Question

Determine the value of the equilibrium constant, Kgoal, for the reaction

N2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=?

by making use of the following information:

1. N2(g)+O2(g)⇌2NO(g), K1 = 4.10×10−31
2. N2(g)+2H2(g)⇌N2H4(g), K2 = 7.40×10−26
3. 2H2O(g)⇌2H2(g)+O2(g), K3 = 1.06×10−10
Express your answer numerically.

Kgoal =
Part B

Determine the equilibrium constant, Kgoal, for the reaction

4PCl5(g)⇌P4(s)+10Cl2(g), Kgoal=?

by making use of the following information:

P4(s)+6Cl2(g)⇌4PCl3(g), K1=2.00×1019

PCl5(g)⇌PCl3(g)+Cl2(g), K2=1.13×10−2

Express your answer numerically.

Kgoal =

Answer 1

Answer:

Part A

$K_{goal}= 3.22\times 10 ^{-66}$

Part B

$K_{goal}= 2.26\times10^{-21}$

Explanation:

Part A:

The given equation is not balanced. So, initially let's balance the equation by taking 24 moles of each of the reagent and NO.

24N₂ + 24H₂O ⇌ 24NO + 12N₂H₄

Now, simplify the equation by dividing both sides by 12. The final balanced equation is the following

2N₂ + 2H₂O ⇌ 2NO + N₂H₄

The above-balanced equation can be solved algebraically to obtain the required Kgoal value.

Adding given equations 1, 2 and 3 we obtain the required equation.

When the equations are added, the equilibrium constants of each equation are multiplied. Mathematically it can be represented as,

$K_1 \times K_2 \times K_3 = K_{goal}$

$K_{goal}= 4.10 \times 10^{-31} \times 7.40 \times 10^{-26} \times 1.06 \times 10^{-10}$

$K_{goal}= 3.22\times 10 ^{-66}$

Part B:

The required equation is balanced, Now

Let.

P₄(s)+6Cl₂(g) ⇌ 4PCl₃(g), K₁=2.00×10¹⁹ ------------------------------------ (a)

PCl₅(g) ⇌ PCl₃(g)+Cl₂(g), K₂=1.13×10⁻² --------------------------------------- (b)

By multiplying equation 2 by 4 and subtracting equation 1 from it, we get

4PCl₅(g) ⇌ P₄(s)+10Cl₂(g)

The Kgoal for the above equation is the product of four times K₂ and inverse K₁ according to the applied operation. Mathematically,

$K_{goal}= 4K_2 \times \frac{1}{K_1}$

$K_{goal}= 4(1.13\times10^{-2}) \times \frac{1}{2.00\times10^{19}}$

$K_{goal}= 2.26\times10^{-21}$