# chem a level question

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#1
does anyone know how to answer this qestion?

When one mole of ammonia is heated to a high temperature, 50% dissociates according to the following equilibrium.
2NH3(g) ⇌ N2(g) + 3H2(g)

What is the total number of moles of gas present in the equilibrium mixture?

thanks
0
1 month ago
#2
So, think about how many moles of each you had before the eqb was established.

You had 1mol of NH3, and 0 of N2 and 0 of H2

When it reached eqb, 1/2 of the 1mol of NH3 disassociates whilst the other 1/2 remains as 0.5 mol of NH3.

Since there are two moles of NH3, you divide that 0.5 by 2 to find what each mol is, if that makes sense?? So, 0.5/2 = 0.25

One mole of N2 ----> 1 * 0.25 = 0.25 mol
3 mole of H2 -------> 3 * 0.25 = 0.75

0.75 + 0.25 + 0.5 = 1.5 mol all together.

I hope this made sense (I'm sorry if my explanation is a bit dodgy lol)
0
1 month ago
#3
(Original post by he3456)
does anyone know how to answer this qestion?

When one mole of ammonia is heated to a high temperature, 50% dissociates according to the following equilibrium.
2NH3(g) ⇌ N2(g) + 3H2(g)

What is the total number of moles of gas present in the equilibrium mixture?

thanks

If none dissociated, you'd still have 1 mol.

If all dissociated, you'd have 2 mol (since 2 mol of reactant turns into 4 mol of products)

If half of the reactant doesn't dissociate and half does...
0
#4
(Original post by inoubliable)
So, think about how many moles of each you had before the eqb was established.

You had 1mol of NH3, and 0 of N2 and 0 of H2

When it reached eqb, 1/2 of the 1mol of NH3 disassociates whilst the other 1/2 remains as 0.5 mol of NH3.

Since there are two moles of NH3, you divide that 0.5 by 2 to find what each mol is, if that makes sense?? So, 0.5/2 = 0.25

One mole of N2 ----> 1 * 0.25 = 0.25 mol
3 mole of H2 -------> 3 * 0.25 = 0.75

0.75 + 0.25 + 0.5 = 1.5 mol all together.

I hope this made sense (I'm sorry if my explanation is a bit dodgy lol)
ah, thank you, it definitely made sense 1
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