Chemistry/The equilibrium number/ I really

In summary, the conversation is discussing the equilibrium constant and moles of substances in a chemical reaction, specifically the reaction N2 (g) + 3H2 (g) <===> 2NH3 (g). The initial amount of N2 is 2 moles and the initial amount of H2 is 4 moles in a 0.4 L container. At equilibrium, 1.14 moles of N2 remain. The questions involve finding the equilibrium number of moles of hydrogen, the amount of ammonia produced, and the equilibrium constant. These can be solved using the ICE table, where x represents the change in N2, y represents the change in H2, and z represents the change in NH3
  • #1
hedi
1
0
1. N2 (g) + 3H2 (g) <===> 2NH3 (g)
initial 2mol ----- 4mol -------------- 0
change -x ------ -y ---------------- +z
End 1.14

2 mol N2 mixed with 4 mol H2 in 0.4 L container. At equilibrium 1.14 of N2 remain


There were 2 other questions too:

What is Y and what is X. I solved those but I can't do the rest, my answers are wrong

1) What is the equilibrium number of moles of hydrogen?

2) What is z, that is, how much ammonia is produce?

3) What is the equilibrium constant, Keq?



I tried to answer these questions over and over but it turns out wrong. I need hel soon, I'm stuck .please help
 
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  • #2
I am not sure what the quastion is - were you given the ICE table, or have you made it on your own, assigning x, y and z to different reagents?

x, y and z are connected by stoichiometry of the reaction. For example - for every mole of N2 that reacted, 3 moles of NH3 are produced, thus z=3x.

--
methods
 
  • #3


1) The equilibrium number of moles of hydrogen (H2) can be determined using the equilibrium constant expression: Keq = [NH3]^2 / [N2]

^3. We know that at equilibrium, 1.14 moles of N2 remain. Therefore, we can set up the following equation: Keq = (2-x)^2 / (2-x)(4-y)^3. We also know that the initial number of moles of N2 and H2 were 2 and 4, respectively. Therefore, at equilibrium, the number of moles of H2 must be (4-y). Substituting this into the Keq expression, we have: Keq = (2-1.14)^2 / (2-1.14)(4-y)^3. Solving for y, we get a value of approximately 3.384 moles of H2 at equilibrium.

2) The value of z represents the amount of ammonia (NH3) produced at equilibrium. From the equation, we can see that for every 1 mole of N2 that reacts, 2 moles of NH3 are produced. Therefore, since 0.86 moles of N2 reacted (2-1.14), we can expect 1.72 moles of NH3 to be produced at equilibrium.

3) The equilibrium constant, Keq, can be calculated using the following formula: Keq = [NH3]^2 / [N2]

^3. Substituting the values from the equilibrium state, we have: Keq = (1.72)^2 / (1.14)(3.384)^3. Solving for Keq, we get a value of approximately 0.29. This value represents the equilibrium constant for the given reaction at the specified conditions.

 

Related to Chemistry/The equilibrium number/ I really

What is chemistry?

Chemistry is the scientific study of matter, its properties, and the changes it undergoes. It is a central science, as it connects and overlaps with other fields such as biology, physics, and geology.

How is equilibrium number defined?

The equilibrium number is a term used in chemistry to describe the number of molecules or moles of a substance that are present in a reaction mixture at equilibrium. It can also refer to the ratio of the concentrations of products to reactants at equilibrium.

What factors influence the equilibrium number?

The equilibrium number is influenced by several factors, including temperature, pressure, and the concentration of reactants and products. These factors can shift the equilibrium position and change the equilibrium number accordingly.

Why is understanding equilibrium important in chemistry?

Equilibrium is important in chemistry because it helps us predict the direction of a chemical reaction and the amount of product that will be formed. It also allows us to optimize reaction conditions to achieve a desired equilibrium number.

What are some real-life applications of equilibrium in chemistry?

Equilibrium is crucial in many real-life applications, such as in the production of chemicals, drugs, and food products. It is also used in environmental and industrial processes, such as in air and water purification, and in the design of efficient energy storage systems.

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