. For which reaction is īSsys Ĩ 0?
A) CH4(g) + 2 O2(g) ¡÷ CO2(g) + 2 H2O(l)
B) NH3(g) + HCl(g) ¡÷ NH4Cl(s)
C) 2 NH4NO3(s) ¡÷ 2 N2(g) + O2(g) + 4 H2O(g)
D) HCl(aq) + NaOH(aq) ¡÷ NaCl(aq) + H2O(l)
I don't understand the symbols you've used.
To determine which reaction has a standard entropy change (ΔSsys) of 0, we need to consider the stoichiometry of the reactions involved. The standard entropy change (∆Ssys) is related to the difference in entropy between the products and the reactants.
In the given options:
A) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
B) NH3(g) + HCl(g) → NH4Cl(s)
C) 2 NH4NO3(s) → 2 N2(g) + O2(g) + 4 H2O(g)
D) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
To find ΔSsys, we need to consider the change in the number of moles of gaseous species (Δn) in the reaction. The equation for ΔSsys is given by the formula:
ΔSsys = Σn(prod)S(prod) - Σn(react)S(react)
Where Σn is the sum of the stoichiometric coefficients, and S is the molar entropy.
Looking at the options, we can evaluate the value of Δn for each reaction:
A) Δn = (1 + 2) - (1 + 2) = 0
B) Δn = (1 + 1) - 1 = 1
C) Δn = (2 + 1 + 4) - 2 = 5
D) Δn = (1 + 1) - 1 = 1
We see that option A has a value of Δn equal to 0, which indicates that there is no change in the number of moles of gaseous species. Therefore, the reaction A) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) has a ∆Ssys of 0.