Answer : The value of equilibrium constant for this reaction at 328.0 K is 
Explanation :
As we know that,
where,
= standard Gibbs free energy = ?
= standard enthalpy = 151.2 kJ = 151200 J
= standard entropy = 169.4 J/K
T = temperature of reaction = 328.0 K
Now put all the given values in the above formula, we get:
The relation between the equilibrium constant and standard Gibbs free energy is:
where,
= standard Gibbs free energy = 95636.8 J
R = gas constant = 8.314 J/K.mol
T = temperature = 328.0 K
K = equilibrium constant = ?
Now put all the given values in the above formula, we get:
Therefore, the value of equilibrium constant for this reaction at 328.0 K is
Matter is every object that stands on this Earth
Answer:
A decrease in [H3O+] and an increase in pH (option a)
Explanation:
Equilibrium of water is shown in this equation
2H₂O ⇄ H₃O⁺ + OH⁻
When you add NaOH, you are modifying [OH⁻]
NaOH → Na⁺ + OH⁻
In equilibrium of water, the [OH⁻] increases
2H₂O ⇄ ↓ H₃O⁺ + OH⁻ ↑
As the [OH⁻] increases, by Le Chatellier, the equilibrium tends to decrease [H₃O⁺].
If the [OH⁻] is higher, pH is also high so the solution of water and sodium hydroxide would be totally basic.
1 mole = 6.22 x 10^23 molecules (Avogadro's number)
15 moles x (6.22 x 10^23) = 9.33 x 10^24 atoms
