Consider the reaction SnO2(s) + 2C(graphite) + 2Cl2(g) SnCl4(l) + 2CO(g) Determine the following at 298 K: Ho = -186.13 Correct: Your answer is correct. kJ/mol So = 144.75 Correct: Your answer is correct. J/mol-K Go = 229.04 Incorrect: Your answer is incorrect. kJ/mol What is log K at 298K (log K = ln K/2.3026)? log K = 40.15 Correct: Your answer is correct. (Hint log K = (ln K)/2.3026) Assume that Ho and So are temperature independent to determine the following at 800 K. Go = kJ/mol log K = (log K = ln K/2.3026)
Consider the reaction SnO2(s) + 2C(graphite) + 2Cl2(g) SnCl4(l) + 2CO(g) Determine the following at 298...
macnillan learning Consider the following reaction at 298 K: 2C(graphite +01g) -> 2CO(g) F1° =-221.0 kJ / mol Calculate the following quantities. Find standard entropy values here. Number J/ (mol K sys Number J/(mol K surr Number Previous ⓧ Gve Up & View Solution > Check Answer Next Exit ..
Consider the reaction Al2O3(s) + 3C(graphite) + 3Cl2(g) → 2AICI3(s) + 3CO(g) Determine the following at 298 K: X kJ/mol J/mol-K kJ/mol дно- What is log K at 298K (log K-In K/2.3026)? log K (Hint log K-(In K)/2.3026) Assume that ΔΗο and ΔSo are temperature independent to determine the following at 600 K. kJ/mol log K (log K = In K/2.3026)
Consider the following reaction at 298 K. C(graphite)+2Cl2(g)⟶CCl4(l)ΔH∘=−139 kJC(graphite)+2Cl2(g)⟶CCl4(l)ΔH∘=−139 kJ Calculate the following quantities. Refer to the standard entropy values as needed. Consider the following reaction at 298 K. C(graphite) + 2Cl2(g) C014 (1) AH = -139 kJ Calculate the following quantities. Refer to the standard entropy values as needed. A$sys = 179.42 ASsurr = 466.44 A.Sunix =
and 50. Given the following, determine AGof at 298 K for SnO. Sn(s)+SnO2(s) 2SnO(s); AG° = 12.0 kJ at 298K Substance AGOF(kJ/mol) at 298 K SnO(s) SnO2(s) ? -515.8 95. Consider the following reaction: 3C(s)+4H2(g) -C3H8(g); AH =-104.7 kJ; AS° = -287.4J/K at 298 K What is the equilibrium constant at 400.0 K for this reaction?
For the reaction S(s,rhombic) + 2CO(g) SO2(g) + 2C(s,graphite) AH = -75.8 kJ and AS =-167.6J/K The standard free energy change for the reaction of 1.88 moles of S(s,rhombic) at 294 K, 1 atm would be This reaction is (reactant, product) favored under standard conditions at 294 K. Assume that AH° and AS are independent of temperature. Submit Answer Try Another Version 2 Item attempts remaining
The standard enthalpy change for the following reaction is -111 kJ at 298 K. C(s,graphite) + 1/2 O2(g) --> CO(g) ΔH° = -111 kJ What is the standard enthalpy change for the reaction at 298 K? 2 CO(g) --> 2 C(s,graphite) + O2(g) Answer in kJ
4. Consider the reaction 3 C (graphite)2 H2O (g) CH4 (g) 2 CO (g) The following data are needed (all are at 298 K): S(J mol K) AH(kJ mol AAG(kJ mol) compound C, graphite 0 5.740 Н.О (g) - 241.82 188.83 - 228.57 CH4 (g) 186.26 -74.81 - 50.72 СО (g) 197.67 - 137.17 - 110.53 a) Determine A,G° and K for this reaction at 298 K and comment on the spontaneity of this process b) If you find that...
Consider the following reaction at 298 K.C(graphite) + 2H2(g)→ CH4(g) ΔH°=-74.6 kJ
Using the following data determine the temperature (in K) at which the reaction H2O(g)+ C(s,graphite) ↔ H2(g) + CO(g) this becomes spontaneous. ΔfH° (H2O(g)) = -251.2 kJ mol-1 ΔfH° (C(s,graphite)) = 0.0 kJ mol-1 ΔfH° (H2(g)) = 0.0 kJ mol-1 ΔfH° (CO(g)) = -110.1 kJ mol-1 S° (H2O(g)) = 192.6 J K-1 mol-1 S° (C(s,graphite)) = 6.4 J K-1 mol-1 S° (H2(g)) = 136.9 J K-1 mol-1 S° (CO(g)) = 192.4 J K-1 mol-1
Consider the following reaction at 298 K. C(graphite)+2H2(g)⟶CH4(g)ΔH∘=−74.6 kJ and ΔS∘=−80.8 J/KC(graphite)+2H2(g)⟶CH4(g)ΔH∘=−74.6 kJ and ΔS∘=−80.8 J/K Calculate the following quantities. ΔSsys=ΔSsys J/K ΔSsurr= J/K ΔSuniv= J/K