The given chemical reaxction is as follows:
2NiS(s) + 3O2(g) -> 2SO2(g) + 2NiO(s)
The standard entropy for the reaction is
Δ S rxn = Σn*ΔS0 products - Σn*ΔS0 reactants (where n,m are the number of moles)
=[ 2*ΔS0 SO2(g) + 2*ΔS0 NiO(s) ]- [ 2*ΔS0 NiS() + 3*ΔS0O2(g) ]
= [2*248 J/K + 2* 38.0 J/K ]-[2*53.0J/K + 3*205J/K]
= [496J/K + 76J/K ]-[106J/K + 615J/K]
= 572 J/K - 721 J/K
= -149J/K
Therefore Δ S rxn = -149J/K
Calculate the standard-state entropy for the following reaction: 2NiS(s) + 3O2(g) -> 2SO2(g) + 2NiO(s) The sta...
1) Calculate the standard-state entropy for the following reaction:2 NiS(s)+ 3 O2(g)→ 2 SO2(g)+ 2 NiO(s)(If applicable, coefficients of one have been included for clarity.)The standard entropy values are given in the table.Formula S J/(K*mol)SO2(g) 248NiO(s) 38.0NiS(s) 53.0O2(g) 205Express your answer in joules per kelvin to three significant figures.
Use the given data at 500 K to calculate ΔG°for the reaction 2H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g) Substance H2S(g) O2(g) H2O(g) SO2(g) ΔH°f(kJ/mol) -21 0 -242 -296.8 S°(J/K·mol) 206 205 189 248
Calculate the standard-state entropy for the following reaction: 6 CO2(g) + 6 H2O(l) ⟶ 1 C6H12O6(s) + 6 O2(g) (If applicable, coefficients of one have been included for clarity.) The standard entropy values are given in the table. Formula S∘ J/(K⋅mol) C6H12O6(s) 212 O2(g) 205 CO2(g) 214 H2O(l) 189
Calculate ΔS∘rxn for the following balanced chemical equation: 2H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g) Substance and state S∘[J/(K⋅mol)] H2O(g) 188.8 O2(g) 205 H2S(g) 205.7 SO2(g) 248.1
Calculate the standard-state entropy for the following reaction: 1 CH4(g) + 2 O2(g) ⟶ 1 CO2(g) + 2 H2O(l) (If applicable, coefficients of one have been included for clarity.) The standard entropy values are given in the table. Formula S∘ J/(K⋅mol) CO2(g) 214 H2O(l) 189 CH4(g) 186 O2(g) 205
Consider the reaction: 2SO2(g) + O2(g)2SO3(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 2.11 moles of SO2(g) react at standard conditions. S°system = J/K
Consider the reaction 2SO2(g) + O2(g)2SO3(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.11 moles of SO2(g) react at standard conditions. S°surroundings = J/K
For the following reaction: CS2(g) + 3O2(g) → CO2(g) + 2SO2(g) Compound ΔH°f (kJ mol-1) S° (J mol-1 K-1) CS2 (g) 116.70 237.80 O2 (g) 0.00 205.70 CO2 (g) -393.51 213.74 SO2 (g) -296.84 248.20 Calculate ΔG°rx (in kJ) at 794.8 K for this reaction. Report your answer to two decimal places in standard notation (i.e. 123.45 kJ). Assume ΔH°f and S° do not vary as a function of temperature.
Consider the following reaction at 298 K. 4Al(s)+3O2(g)⟶2Al2O3(s) Δ?∘=−3351.4 kJ Calculate the following quantities. Refer to the standard entropy values as needed. Δ?sys= J/K Δ?surr= J/K Δ?univ= J/K Al Entropy = 28.3 O2 Entropy=205.2 Al2O3= 50.9
Calculate the standard-state entropy for the following reaction: 1 Al2O3(s) + 3 H2(g) ⟶ 2 Al(s) + 3 H2O(l) (If applicable, coefficients of one have been included for clarity.) The standard entropy values are given in the table. Formula S∘ J/(K⋅mol) Al(s) 28.0 H2O(l) 189 Al2O3(s) 51.0 H2(g) 131