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.
Explanation:
sorry po king Hindi kopo ma sagot
What pH scale number represents a neutral.
The answer is 7
Answer:
The concentrations of A, B, and C at equilibrium:
[A] = 0.0 M
[B] = 2.7 M
[C] = 2.4 M
Explanation:
Concentration of 1.80 mol of A in 1.00 L container :
![[A]=\frac{1.80 mol}{1.00 L}=1.80 M](https://tex.z-dn.net/?f=%5BA%5D%3D%5Cfrac%7B1.80%20mol%7D%7B1.00%20L%7D%3D1.80%20M)
Concentration of 3.90 mol of B in 1.00 L container :
![[B]=\frac{3.90 mol}{1.00 L}=3.90 M](https://tex.z-dn.net/?f=%5BB%5D%3D%5Cfrac%7B3.90%20mol%7D%7B1.00%20L%7D%3D3.90%20M)
Initially
1.80 M 3.90 M 0
At equilibrium
(1.80-3x)M (3.90-2x) 4x
The expression of an equilibrium constant is given by :
![K_c=\frac{[C]^4}{[A]^3[B]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%5E4%7D%7B%5BA%5D%5E3%5BB%5D%5E2%7D)
Solving for x:
x = 0.600
The concentrations of A, B, and C at equilibrium:
[A] = [1.80-3x]=[1.80-3 × 0.600]= 0 M
[B] = [3.90-2x] = [3.90-2 × 0.600] = 2.7 M
[C] = [4x] =[4 × 0.600 M] = 2.4 M
