Consider the following reaction at constant P. Use the information here to determine the value of ΔSsurr at 398 K. Predict whether or not this reaction will be spontaneous at this temperature.
4 NH3(g) + 3 O2(g) → 2 N2(g) + 6 H2O(g); ΔH = -1267 kJ
A)ΔSsurr = +3.18 kJ/K, reaction is spontaneous
B)ΔSsurr = -12.67 kJ/K, reaction is spontaneous
C)ΔSsurr = +12.67 kJ/K, reaction is not spontaneous
D)ΔSsurr = +50.4 kJ/K, reaction is not spontaneous
E)ΔSsurr = -3.18 kJ/K, it is not possible to predict the spontaneity of this reaction without more information.
To determine the value of ΔSsurr (change in entropy of the surroundings), we first need to calculate ΔSsys (change in entropy of the system).
The formula for ΔSsys is given by ΔSsys = ΣnS(products) - ΣnS(reactants), where Σn represents the stoichiometric coefficients and S denotes the molar entropy of each component.
For the given reaction:
4 NH3(g) + 3 O2(g) → 2 N2(g) + 6 H2O(g)
We can use the molar entropies given in a reference table to calculate ΔSsys = ΣnS(products) - ΣnS(reactants).
Now, to determine whether the reaction is spontaneous at 398 K, we can use the equation ΔG = ΔH - TΔS, where ΔH is the enthalpy change, T is the temperature in Kelvin, and ΔS is the entropy change.
If ΔG < 0, the reaction is spontaneous. If ΔG > 0, the reaction is non-spontaneous.
Let's calculate ΔSsys and then evaluate the spontaneity of the reaction:
ΔSsys = (2 mol * S(N2) + 6 mol * S(H2O)) - (4 mol * S(NH3) + 3 mol * S(O2))
Given the values of molar entropies at 398 K:
S(NH3) = 0.192 J/(mol*K)
S(O2) = 0.205 J/(mol*K)
S(N2) = 0.192 J/(mol*K)
S(H2O) = 0.188 J/(mol*K)
Now, substitute the values and calculate ΔSsys.
ΔSsys = (2 * 0.192 + 6 * 0.188) - (4 * 0.192 + 3 * 0.205)
= 1.664 - 1.096
= 0.568 J/K
Now, to determine ΔSsurr, we can use the equation ΔSsurr = -ΔHsys / T.
Given ΔH = -1267 kJ = -1267000 J, and T = 398 K.
ΔSsurr = -ΔHsys / T
= -(-1267000) / 398
= 3186.43 J/K
≈ +3.19 kJ/K
Now that we have found ΔSsurr to be approximately +3.19 kJ/K, we can compare it to zero. If ΔSsurr > 0, it indicates that the surroundings are becoming more disordered, which favors the spontaneity of the reaction.
Therefore, based on the positive value of ΔSsurr, the reaction will be spontaneous at 398 K.
The correct answer is A) ΔSsurr = +3.18 kJ/K, and the reaction is spontaneous.
To determine the value of ΔSsurr at 398 K, we need to calculate the ΔSsys (change in entropy of the system) and then use the fact that ΔSuniv = ΔSsys + ΔSsurr = 0 for a system at constant pressure.
To find ΔSsys, we use the equation:
ΔSsys = ΣnS(products) - ΣnS(reactants)
where n is the stoichiometric coefficient and S is the molar entropy.
From a table of molar entropies, we find:
S(NH3(g)) = 192.77 J/(mol·K)
S(O2(g)) = 205.15 J/(mol·K)
S(N2(g)) = 191.61 J/(mol·K)
S(H2O(g)) = 188.72 J/(mol·K)
Plugging these values into the equation, we have:
ΔSsys = (2×191.61 J/(mol·K) + 6×188.72 J/(mol·K)) - (4×192.77 J/(mol·K) + 3×205.15 J/(mol·K))
ΔSsys = 384.33 J/(mol·K) - 1003.16 J/(mol·K)
ΔSsys = -618.83 J/(mol·K)
Now, using the fact that ΔSuniv = ΔSsys + ΔSsurr = 0, we can find the value of ΔSsurr:
ΔSsurr = -ΔSsys
ΔSsurr = -(-618.83 J/(mol·K))
ΔSsurr = 618.83 J/(mol·K)
Since ΔSsurr is positive, the correct answer is:
A) ΔSsurr = +3.18 kJ/K, and the reaction is spontaneous.