# Equilibrium Constant Keq: 2H2S(g) <-> H2(g) + S2(g) at 830°C

• Fusilli_Jerry89
In summary, the value of Keq for the reaction 2H2S(g) <-> 2H2(g) + S2(g) is 4.20x10^-6 at 830 degrees celsius. The products are favoured in this reaction, as determined by the rules of equilibrium. To solve for the concentration of S2 at equilibrium after injecting 0.200 mol of H2S into a 1.00 L flask, an ice box can be used. This results in an equation of 4x^3 - 1.68x10^-5x^2 + 3.36x10^-6 - 1.68x10^-7 = 0, which can be solved for
Fusilli_Jerry89
The value of Keq for the reaction: 2H2S(g) <-> 2H2(g) + S2(g) is 4.20x10^-6 at 830 degrees celsius.
a) What is favoured in this reaction, reactants or products?
b) What concentration of S2 can be expected at equilibrium after 0.200 mol of H2S is injected into an empty 1.00 L flask?

a) Do you know how to write out Keq? (Hint: Use the law of mass action).
b) Make an ice box
2H2S(g) <--> 2H2(g) + S2 (g)
i .200 0 0
c -2x +2x +x
e .200 - 2x 2x x

Keq = [((2x)^2)(x)] / [(.200 - 2x)^2]

Last edited:
i solved x to equal 1.431 but when u plug it back into the original question, H2S comes out to a negative number

oh nm i made a mistake

k got down to 4x^3-1.68x10^-5x2+3.36x10^-6-1.68x10^-7=0 now what

Fusilli_Jerry89 said:
k got down to 4x^3-1.68x10^-5x2+3.36x10^-6-1.68x10^-7=0 now what
What does that mean?

Look to your rules of equilibrium to determine which side of the reaction is favoured.

As for x, you can solve for it.

i put the products are favoured, but i do not know how to solve this equation because it has an 4^3 x^2 and an x

Fusilli_Jerry89 said:
i put the products are favoured, but i do not know how to solve this equation because it has an 4^3 x^2 and an x
Why do you have $4x^{3}$ in your equation? Solve for x.

## 1. What is the equilibrium constant (Keq) for the reaction 2H2S(g) <-> H2(g) + S2(g) at 830°C?

The equilibrium constant (Keq) for this reaction is equal to the ratio of the products to the reactants when the reaction reaches equilibrium at 830°C. In this case, the Keq is equal to [H2][S2]/[H2S]^2.

## 2. How does the temperature affect the equilibrium constant (Keq)?

The equilibrium constant (Keq) is temperature-dependent. As the temperature increases, the Keq value also increases, indicating a greater amount of product formed at equilibrium. In this reaction, the Keq value at 830°C will be different from the Keq value at a different temperature.

## 3. What does a high or low equilibrium constant (Keq) value indicate?

A high Keq value indicates that at equilibrium, there is a higher concentration of products compared to reactants, meaning that the reaction favors the formation of products. A low Keq value indicates that at equilibrium, there is a higher concentration of reactants compared to products, meaning that the reaction favors the formation of reactants.

## 4. Can the equilibrium constant (Keq) be changed?

The equilibrium constant (Keq) is a constant value for a given reaction at a specific temperature. It cannot be changed but can be influenced by changing the temperature, pressure, or concentrations of reactants or products.

## 5. How is the equilibrium constant (Keq) related to the direction of the reaction?

The equilibrium constant (Keq) is related to the direction of the reaction by indicating which side of the reaction equation is favored at equilibrium. A Keq value greater than 1 indicates that the reaction favors the forward direction, while a Keq value less than 1 indicates that the reaction favors the reverse direction.

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