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4vir4ik [10]
3 years ago
10

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

Chemistry
1 answer:
hodyreva [135]3 years ago
5 0

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}

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