If .75 moles of ammonia is needed, how many grams of nitrogen will be consumed?
1 answer:
We have to get the amount of nitrogen to be consumed to get 0.75 moles of ammonia.
The amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is: 10.5 grams.
Ammonia (NH₃) can be prepared from nitrogen (N₂) as per following balanced chemical reaction-
N₂ (g) + 3H₂ (g) ⇄ 2NH₃ (g)
According to the above reaction, to prepare 2 moles of ammonia, one mole of nitrogen is required. Hence, to prepare 0.75 moles of ammonia, moles = 0.375 moles of nitrogen is required.
Molar mass of nitrogen is 28 grams, i.e, mass of one mole of nitrogen is 28 grams, so mass of 0.375 moles of nitrogen is 0.375 X 28 grams=10.5 grams of nitrogen.
Therefore, the amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is 10.5 grams.
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Answer:
Explanation:
From the given information;
The chemical reaction can be well presented as follows:
⇄
Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:
i.e
However, since we are dealing with liquids solutions;
since the activity of
is equivalent to 1
Hence, under standard conditions(i.e at a pressure of 1 bar)
(b)
From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:
Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol
(c)
Le's recall that:
At equilibrium, the instantaneous free energy is usually zero &
Q(reaction quotient) is equivalent to K(equilibrium constant)
So;
(d)
The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).
This is because;
If Q < K, then the reaction will proceed in the right direction towards the products.
However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.
So;
Since we are dealing with liquids;
Q = 1
Since Q < K; Then, the reaction proceeds in the right direction.
Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.