The equilibrium constant of a reaction is defined as:
"The ratio between equilibrium concentrations of products powered to their reaction quotient and equilibrium concentration of reactants powered to thier reaction quotient".
The reaction quotient, Q, has the same algebraic expressions but use the actual concentrations of reactants.
To solve this question we need this additional information:
<em>For this reaction, K = 6.0x10⁻² and the initial concentrations of the reactants are:</em>
<em>[N₂] = 4.0M; [NH₃] = 1.0x10⁻⁴M and [H₂] = 1.0x10⁻²M</em>
<em />
Thus, for the reaction:
N₂ + 3H₂ ⇄ 2NH₃
The equilibrium constant, K, of this reaction, is defined as:
![K = 6.0x10^{-2} = \frac{[NH_3]^2}{[N_2][H_2]^3}](https://tex.z-dn.net/?f=K%20%3D%206.0x10%5E%7B-2%7D%20%3D%20%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_2%5D%5E3%7D)
And Q, is:
![Q = \frac{[NH_3]^2}{[N_2][H_2]^3}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_2%5D%5E3%7D)
Where actual concentrations are:
[NH₃] = 1.0x10⁻⁴M
[N₂] = 4.0M
[H₂] = 2.5x10⁻¹M
Replacing:
![Q = \frac{[1.0x10^{-4}]^2}{[4.0][2.5x10^{-1}]^3}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B%5B1.0x10%5E%7B-4%7D%5D%5E2%7D%7B%5B4.0%5D%5B2.5x10%5E%7B-1%7D%5D%5E3%7D)
<h3>Q = 1.6x10⁻⁷</h3>
As Q < K,
<h3>The chemical system will shift to the right in order to produce more NH₃</h3>
Learn more about chemical equililbrium in:
brainly.com/question/24301138
ΔS =S(products) -S(reactants)
Where ΔS is the change of entropy in a reactions
a. ΔS = (2) - (2+1) = -1
b. ΔS = (1+1) -(1) = 1
c. ΔS = (1+2) - (1) = 2
d. ΔS = (2) - (2+1) = -1
e. ΔS = (1) - (1) = 0
ΔS is negative for reaction a. and d.
Answer:
Percent loss of water = 25%
Explanation:
Given data:
Mass of hydrated salt = 15.6 g
Mass of anhydrous salt = 11.7 g
Percentage of water lost = ?
Solution:
First of all we will calculate the mass of water in hydrated salt.
Mass of water = Mass of hydrated salt - Mass of anhydrous salt
Mass of water = 15.6 g - 11.7 g
Mass of water = 3.9 g
Now we will calculate the percentage.
Percent loss of water = mass of water / total mass × 100
Percent loss of water = 3.9 g/ 15.6 g × 100
Percent loss of water = 25%
The answer to this is A i think.
Answer:It is a mixture
Explanation:
If it is pure sugar it’s neither but if it has water it is a homogeneous mixture