For the reaction below at 298.15 K, the standard enthalpy of reaction is ΔrH° = 41.6 kJ mol−1 and the thermodynamic equilibrium constant is K = 1.50×10−6.
2 NO(g) + I2(s) ⇌ 2 NOI(g)
What is ΔrS° for this reaction?
Do I need to take the K value into consideration or can I just solve using n and S values?
I think you must use K.
dGo = -RT*lnK
Substitute and solve for dGo.
Then dGo = dHo - TdSo and solve for dSo.
dGo = -RT*lnK
for this would it be -8.135(298)ln(298)?
No, You have substituted 298 for k but that isn't k. The other substitutions look OK to me.
what would k be then?
C'mon. K is given to you in the problem.
To determine the standard entropy change (ΔrS°) for a reaction, you can use the equation:
ΔrG° = ΔrH° - TΔrS°
Where:
ΔrG° = standard Gibbs free energy change
ΔrH° = standard enthalpy change
T = temperature in Kelvin
ΔrS° = standard entropy change
Given that ΔrG° can be calculated from the equilibrium constant K using the equation:
ΔrG° = -RT ln(K)
Where:
R = gas constant (8.314 J/(mol·K))
T = temperature in Kelvin
ln = natural logarithm
K = equilibrium constant
You can see that the value of K is already provided in the question (K = 1.50×10−6). Since ΔrG° and ΔrH° can be related using the equation above, knowing the value of K is necessary to get the ΔrS°.
Therefore, you will need to consider the value of K in order to calculate the ΔrS° for this reaction.