Use the following information to answer the next question

Reaction 1: 2X(g) + 3Y(g) = X_{2}*Y_{3}(g) K = 0.25

Reaction 2: X 3 Y 3 (g) rightleftharpoons 2X(g) + 3Y(g)

K=?

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2. The unknown value of K in Reaction 2 will be

Given that Reaction 1 and Reaction 2 are the reverse of each other, we can use the relationship between the equilibrium constants of forward and reverse reactions.

For Reaction 1:
\[ 2X(g) + 3Y(g) \rightleftharpoons X_2Y_3(g) \]
\[ K_1 = 0.25 \]

For Reaction 2, which is the reverse reaction:
\[ X_2Y_3(g) \rightleftharpoons 2X(g) + 3Y(g) \]
\[ K_2 = ? \]

The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction:

\[ K_2 = \frac{1}{K_1} \]

Substituting the given value of \( K_1 \):

\[ K_2 = \frac{1}{0.25} \]
\[ K_2 = 4 \]

Therefore, the unknown value of \( K \) in Reaction 2 is \( 4 \).

Do you want details or have any questions?

Here are five questions you can ask:
1. How do you calculate the equilibrium constant for a reaction?
2. What is the significance of the equilibrium constant value?
3. How do forward and reverse reactions relate to each other?
4. Can you explain how to determine the equilibrium constant for a complex reaction?
5. How do changes in temperature affect the equilibrium constant?

**Tip:** Always remember that the equilibrium constant for a reverse reaction is the reciprocal of the equilibrium constant for the forward reaction.

Learn how to calculate the equilibrium constant for a reversible reaction using the relationship between forward and reverse reactions. This problem involves understanding chemical equilibrium, equilibrium constants, and reversible reactions. Suitable for advanced high school students and anyone studying chemistry.

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