a. 2Na Al 03
Equation must have the same number of each element on both sides of the equation.
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 
Answer:
1. The dye that absorbs at 530 nm.
Explanation:
The dye will absorb light to promote the transition of an electron from the HOMO to the LUMO orbital.
The higher the gap, the higher the energy of transition. The energy can be calculated by E = hc/λ, in which h and c are constants and λ is the wavelength.
The equation shows that the higher the energy, the higher the gap and the lower the wavelength.
Therefore, the dye with absorption at 530 nm has the higher HOMO-LUMO gap.
I think the answer is : 1