Answer:
V₂ = 22.84 L
Explanation:
Given data:
Initial volume = 20.0 L
Initial pressure = 1.50 atm
Initial temperature = 23 °C (23 +273 = 296 K)
Final temperature = 271°C (271+273 = 544 K)
Final pressure = 3.50 atm
Final volume = ?
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 1.50 atm × 29.0L ×544 K / 296 K × 3.50 atm
V₂ = 23664 atm .L. K / 1036 atm.K
V₂ = 22.84 L
Answer:

Explanation:
Hello,
In this case, given the amounts of water and carbon dioxide we should invert the given reaction as hydrogen will be producted rather than consumed:

Consequently, the equilibrium constant is also inverted:

In such a way, we can now propose the law of mass action:
![Kc'=\frac{[H_2][CO_2]}{[H_2O][CO]}](https://tex.z-dn.net/?f=Kc%27%3D%5Cfrac%7B%5BH_2%5D%5BCO_2%5D%7D%7B%5BH_2O%5D%5BCO%5D%7D)
And we can express it in terms of the initial concentrations of the reactants and the change
due to the reaction extent:
![Kc'=\frac{(x)(x)}{([H_2O]_0-x)([CO]_0-x)}=1.87](https://tex.z-dn.net/?f=Kc%27%3D%5Cfrac%7B%28x%29%28x%29%7D%7B%28%5BH_2O%5D_0-x%29%28%5BCO%5D_0-x%29%7D%3D1.87)
Thus, we compute the initial concentration which are same, since equal amount of moles are given:
![[H_2O]_0=[CO]_0=\frac{0.680mol}{70.0L}=0.0097M](https://tex.z-dn.net/?f=%5BH_2O%5D_0%3D%5BCO%5D_0%3D%5Cfrac%7B0.680mol%7D%7B70.0L%7D%3D0.0097M)
Hence, solving for
by using the quardratic equation or solver, we obtain:

For which the correct value is 0.00561M since the other one will produce negative concentrations of water and carbon monoxide at equilibrium. Therefore, the number of moles of hydrogen at equilibrium for the same 70.0-L container turn out:

Best regards.
Answer C) 0.830 moles
Please disgard what I am saying right now, I can't post since it needs 20 characters atleast. I hope this helped! :3
Answer: The expression for equilibrium constant is ![\frac{[NH_3]^2}{[H_2]^3[N_2]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BH_2%5D%5E3%5BN_2%5D%7D)
Explanation: Equilibrium constant is the expression which relates the concentration of products and reactants preset at equilibrium at constant temperature. It is represented as 
For a general reaction:

The equilibrium constant is written as:
![k_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=k_c%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Chemical reaction for the formation of ammonia is:


Expression for
is:
![k_c=\frac{[NH_3]^2}{[H_2]^3[N_2]}](https://tex.z-dn.net/?f=k_c%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BH_2%5D%5E3%5BN_2%5D%7D)
![1.6\times 10^2=\frac{[NH_3]^2}{[H_2]^3[N_2]}](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E2%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BH_2%5D%5E3%5BN_2%5D%7D)