Answer:
Partial pressure of of CO₂ in the product mixture is 0,20atm
Explanation:
The balance equation is:
2CO(g) + O₂(g) → 2CO₂(g)
Total pressure of CO(g) and O₂(g) gases before reaction at 100,0°C and 1,0L is 1,50 atm. You can say:
X₁ + Y₁ = 1,50atm <em>(1)</em>
Where X₁ is initial partial pressure CO and Y₁ is initial partial pressure of O₂
After reaction partial pressures are:
X₂ = X₁ - 2n = 0; <em>2n = X₁</em>
Y₂ = Y₁ - n
Z₂ = 2n
Where Z₂ is final partial pressure of CO₂
After reaction pressure at 100,0°C and 1,0L is 1,40 atm, that means:
1,40 atm = (Y₂ + Z₂)
1,40 atm = Y₁ - n + 2n
1,40atm = Y₁ + n
1,40 atm = Y₁ + X₁/2 <em>(2)</em>
Replacing (1) in (2)
1,40 atm = 1,50atm - X₁ + X₁/2
-0,10 atm = - X₁/2
<em>0,20 atm = X₁</em>.
As 2n = X₁; 2n =<em> Z₂ = 0,20 atm</em>
I hope it helps!
Answer:
I. Changing the pressure:
Increasing the pressure: the amount of H₂S(g) will increase.
Decreasing the pressure: the amount of H₂S(g) will decrease.
II. Changing the temperature:
Increasing the temperature: the amount of H₂S(g) will decrease.
Decreasing the temperature: the amount of H₂S(g) will increase.
III. Changing the H₂ concentration:
Increasing the H₂ concentration: the amount of H₂S(g) will increase.
Decreasing the H₂ concentration: the amount of H₂S(g) will decrease.
Explanation:
Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
I. Changing the pressure:
When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
For the reaction: CH₄(g) + 2H₂S(g) ⇄ CS₂(g) + 4H₂(g),
The reactants side (left) has 3.0 moles of gases and the products side (right) has 5.0 moles of gases.
Increasing the pressure: will shift the reaction to the side with lower moles of gas (left side), amount of H₂S(g) will increase.
Decreasing the pressure: will shift the reaction to the side with lower moles of gas (right side), amount of H₂S(g) will decrease.
II. Changing the temperature
The reaction is endothermic since the sign of ΔH is positive.
So the reaction can be represented as:
CH₄(g) + 2H₂S(g) + heat ⇄ CS₂(g) + 4H₂(g).
Increasing the temperature:
The T is a part of the reactants, increasing the T increases the amount of the reactants. So, the reaction will be shifted to the right to suppress the effect of increasing T and the amount of H₂S(g) will decrease.
Decreasing the temperature:
The T is a part of the reactants, increasing the T decreases the amount of the reactants. So, the reaction will be shifted to the left to suppress the effect of decreasing T and the amount of H₂S(g) will increase.
III. Changing the H₂ concentration:
H₂ is a part of the products.
Increasing the H₂ concentration:
H₂ is a part of the products, increasing H₂ increases the amount of the products. So, the reaction will be shifted to the left to suppress the effect of increasing H₂ and the amount of H₂S(g) will increase.
Decreasing the H₂ concentration:
H₂ is a part of the products, decreasing H₂ decreases the amount of the products. So, the reaction will be shifted to the right to suppress the effect of decreasing H₂ and the amount of H₂S(g) will decrease.
physical
It can be chemical but it's usually physical as the molecular structure doesn't change
Answer:
copper(ll) has 4 oxygen atoms
I have to see the simulation
