Chemistry/The equilibrium number/ I really

[hedi]

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

 

[Borek]

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.

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methods

 

[Blueshift5]

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.