ΔG° at 450. K is -198.86kJ/mol
The following is the relationship between ΔG°, ΔH, and ΔS°:
ΔH-T ΔS = ΔG
where ΔG represents the common Gibbs free energy.
the enthalpy change, ΔH
The temperature in kelvin is T.
Entropy change is ΔS.
ΔG° = -206 kJ/mol
ΔH° equals -220 kJ/mol
T = 298 K
Using the formula, we obtain:
-220kJ/mol -T ΔS° = -206kJ/mol
220 kJ/mol +206 kJ/mol =T ΔS°.
-T ΔS = 14 kJ/mol
for ΔS-14/298
ΔS=0.047 kJ/mol.K
450K for the temperature Completing a formula with values
ΔG° = (450K)(-0.047kJ/mol)-220kJ/mol
ΔG° = -220 kJ/mol + 21.14 kJ/mol.
ΔG°=198.86 kJ/mol
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Answer:
the correct answer is Metallic bonding
Explanation:
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Answer:
16.9g
Explanation:Cu+2AgNO3→2Ag+Cu(NO3)2
Cu will likely have a +2 oxidation state. It is higher in the activity series than Ag, so it is a stronger reducing agent and will reduce Ag in a displacement reaction. Then you need to balance the coefficients knowing than NO3 is -1 and Ag is +1.
Then to calculate the theoretical yield you need to compare moles of the reactants:
m(Cu)=5g
M(Cu)=63.55
n(Cu)=5/63.55=0.0787
By comparing coefficients you require twice as much silver: 0.157mol
n(Ag)=0.157
M(Ag)=107.86
m(Ag)=0.157x107.86=16.9g
Hence, the theoretical yield of this reaction would be 16.9g
