Answer:
We get ammonia because the forward and reverse reactions are happening at the same rates.
If 3 mol of
H
2
is mixed in a sealed vessel with 1 mol
N
2
under suitable conditions then they will react to form ammonia
N
H
3
:
N
2
+
3
H
2
→
2
N
H
3
At the start of the reaction the concentration of the
N
2
and
H
2
are high. As soon as some
N
H
3
is formed the reverse reaction will start to occur:
2
N
H
3
→
N
2
+
3
H
2
The rate of the reaction depends on concentration so the forward reaction will be fast at first when the concentration of the reactants is high. It will slow down as their concentration decreases.
By the same reasoning the reverse reaction will be slow at first then increase. These two processes continue until a point is reached when the rates of the forward and reverse reactions are equal.
We now state that the reaction has reached equilibrium which we show by:
N
2
+
3
H
2
⇌
2
N
H
3
It is described as "dynamic" because the forward and reverse reactions are happening at the same time although the concentrations of all the species remain constant.
So although
N
H
3
is constantly breaking down, more is being formed at a constant rate.
In the Haber Process the system is actually not allowed to completely reach equilibrium as the process is continuous, as described in Mukhtar's answer.
Explanation:
2Cu + S = Cu₂S
n(Cu)=m(Cu)/M(Cu)
n(Cu)=12.71/63.55=0.2 mol
n(S)=m(s)/M(S)
n(S)=8.000/32.01=0.25 mol
Cu:S 2:1
0.2 mol : 0.25 mol copper deficiency, sulphur in excess
theoretical mass of copper (I) sulfide
m(Cu₂S)=M(Cu₂S)n(Cu)/2
the percentage yield in the reaction is
w=m'(Cu₂S)/m(Cu₂S)=2m'(Cu₂S)/M(Cu₂S)n(Cu)
w=2*14.72/(159.16*0.2)=0.925 (92.5%)
It is correct, next time re-check your answer and don't second guess yourself. ;3
Answer:
Explanation:
For a general equilibrium
aA +bB ⇔ cC + dD ,
the equilibrium constant is K = [C]^c [D]^d / [A]^a[B]^b.
Our reasoning here should be based on the fact that Q has the same expression as K, but is used when the system is not at equilibrium, and the system will react to make Q = K to attain it ( Le Chatelier´s principle ).
So with this in mind, lets answer this question.
1. False: Q can large or small but is not the value of the equilibrium constant, it will predict the side towards the equilibrium will shift to attain it.
2. False: Given the expression for the equilibrium constant, we know if K is small the concentrations of the reactants will be large compared to the equilibrium concentrations of the products.
3. False: when the value of K is large, the equilibrium concentrations of the products will be large and it will lie on the product side.
4. True: From our previous reasongs this is the true one.
5. False: If K is small, the equilibrium lies on the reactants side.
