The standard molar entropy value of N2 (g) is 191.5 J/K-mol,
of O2 (g) is 205.0 J/K-mol and of NO2 (g) is 240.0 J/K-mol. From these values
we can calculate the Delta S for the reaction N2 (g) + 2 O2 (g) produces 2 NO2 (g) to be
a. – 156.5 J/K
b. 636.5 J/K
c. 15.5 J/K
d. – 121.5 J/K
e. – 83.5 J/K
help
dSrxn = (n*dSo products) - (n*dSo reactants)
how do i set this up?
.........N2 + 2O2 ==> 2NO2
.......191.5..205.0...240
dSrxn = (n*products)-(n*reactants)
dSrxn = (2*240) -[(191.5+2(205)]
right.
To calculate the ΔS (change in entropy) for a reaction, we need to use the standard molar entropy values of the reactants and products. In this case, we have:
Reactants:
N2 (g) with a standard molar entropy of 191.5 J/K-mol
O2 (g) with a standard molar entropy of 205.0 J/K-mol
Products:
2 NO2 (g) with a standard molar entropy of 240.0 J/K-mol
The ΔS for the reaction can be calculated using the equation:
ΔS = ΣnΔS(products) - ΣmΔS(reactants)
where n and m are the stoichiometric coefficients of the products and reactants, respectively. In this case, n = 2 (from the balanced equation) and m = 1 for N2 and 2 for O2.
Now let's calculate the ΔS for the reaction:
ΔS = (2 * 240.0 J/K-mol) - (1 * 191.5 J/K-mol) - (2 * 205.0 J/K-mol)
= 480.0 J/K-mol - 191.5 J/K-mol - 410.0 J/K-mol
= -121.5 J/K-mol
Therefore, the ΔS for the reaction N2 (g) + 2 O2 (g) → 2 NO2 (g) is -121.5 J/K-mol.
Hence, the correct option is d. -121.5 J/K.