11.2 atoms H (also mind helping me with my most recent math question on my account?)
Answer:
Maintaining a high starting-material concentration can render this reaction favorable.
Explanation:
A reaction is <em>favorable</em> when <em>ΔG < 0</em> (<em>exergonic</em>). ΔG depends on the temperature and on the reaction of reactants and products as established in the following expression:
ΔG = ΔG° + R.T.lnQ
where,
ΔG° is the standard Gibbs free energy
R is the ideal gas constant
T is the absolute temperature
Q is the reaction quotient
To make ΔG < 0 when ΔG° > 0 we need to make the term R.T.lnQ < 0. Since T is always positive we need lnQ to be negative, what happens when Q < 1. Q < 1 implies the concentration of reactants being greater than the concentration of products, that is, maintaining a high starting-material concentration will make Q < 1.
Answer:
Explanation:
Hrxn = -245.6-296.8+396+50.0 = -96.4 (kJ/mol)
Since Grxn = Hrxn - T*Srxn, where T = 298.15K, we have:
Srxn = (Hrxn - Grxn)/T = (-96.4+75.2)× 1000÷298.15 = -71.1 (j/K mol)
Since: Srxn = Sf (SO2) + Sf (SOCl2)- Sf (SO3) - Sf (SCl2),
or: -71.1 = 248.1 + Sf (SOCl2) - 256.7-184,
Sf (SOCl2) = 256.7+184-71.1-248.1 = 121.5 (J/mol*k)
In order to find T such that Grxn = Hrxn - T*Srxn >= 0. since Hrxn is negative and Srxn is positive, Grxn will always be less than zero. Therefore there won't be a temperature point at which the reaction is going to be non-spontaenous
