Answer will be as follows-
1. Calculate A Gº for the reaction at 25°C 2 SO2(g) + O2(g) → 2 SO3...
5) For the reaction 2 SO2(g) + O2(g) → 2 SO3(g), if initially P(SO2) = 1.2 atm, P(O2) = 1.8 atm, and P(SO3) = 2.1 atm, calculate AG for this reaction at 25°C. The following data is valid at 25°C: AG° (kJ/mol) SO 300.4 SO3 370.4 A) -140.0 kJ/mol B)-141.3 kJ/mol C)-138.7 kJ/mol D) 1,174.7 kJ/mol E) -137.6 kJ/mol
Calculate Δ G for the reaction 2 SO2 (g) + O2 (g) → 2 SO3 (g) when P of SO2 = 0.500 atm, P of O2 = 0.0100 atm, and P of SO3 = 0.100 atm. The value for Δ G^o for this reaction at 298 K is -141.6 kJ. ΔG = ???? kJ I got -160.9 kj but it is incorrect.
Calculate AG for the reaction 2 SO2 (g) + O2 (g) → 2 SO3 (g) when P of SO2 = 0.500 atm, P of O2 = 0.0100 atm, and P of SO3 = 0.100 atm. The value for AGº for this reaction at 298 K is -141.6 kJ. AG= O * kJ Is this reaction spontaneous or non-spontaneous under these conditions? Spontaneous or non-spontaneous? spontaneous
1).From the following enthalpy changes, S (s) +3/2 O2 (g) 2 SO2 (g) SO3 (g) O2 (g)2 SO3 (g) AH =-395.2 kJ AHo 198.2 kJ Calculate the value of AHo for the reaction by using Hess's law of Heat Summation S(s) O2 (g) SO2 (g) 2) Oxyacetylene torches are fueled by the combustion of acetylene, C2H2. 4 CO2 (g) +2 H20 (g) 2 C2H2 + 5 O2 (g) If the enthalpy change for the reaction is -2511.14 kJ/mol, a) How...
3. Consider the following reaction: 2 SO2 (g) + O2 (g) → 2 SO3 (g) AH.x = -197.6 kJ a. If 285.3 g of SO2 is allowed to react with 158.9 g of O2, what is the limiting reactant and theoretical yield of SO3 in liters if the reaction is performed at 315 K and 50.0 mmHg? How much of each reactant remains at the end of the reaction? [10] LR: SO TY: SO2 remaining: O2 remaining: Imol soa 64.000log...
1.For the reaction at equilibrium 2 SO3↔ 2 SO2 + O2 (∆Horxn= 198 kJ/mol), if we increase the reaction temperature, the equilibrium will (1 point ) * No shift None of the above Question lacks sufficient information Shift to the right 2. For the equilibrium reaction 2 SO2(g) + O2(g) ↔ 2 SO3(g), ∆Horxn = -198 kJ/mol. Which one of these factors would cause the equilibrium constant to increase? (1 point ) * Add a catalyst Decrease the temperature None...
Calculate Kp at 298 K for the reaction SO.(g) + NO2()SO3(g)+NO() SO2(g) SO3(g) NO(g) NO2(g) -300.4 k/mol -370.4 kJ/mol 86.7 kJ/mol 51.8 kJ/mol For the reaction 2NO(g)+02(g)- 2NO2( f iniially P(NO)1.5 atm, PO2)-1.4 atm, and P(NOJ-2.0 atrn, calculate Δ@for this reaction at 25°C. The following data is valid at 25°C: NO NO2 AGe(kJ/mol 86.7 51.8
Use the ΔH°f and ΔH°rxn information provided to calculate ΔH°f for SO3(g): 2 SO2(g) + O2(g) → 2 SO3(g) ΔH°rxn = -198 kJ ΔH°f (kJ/mol) SO2(g) -297
Consider the reaction of NO(g) from its elements. 2 502(g) + O2(g) → 2 SO3(9) Use the thermodynamic data given to determine the following for this reaction: AH°, equal O kJ/mol Asº, equals O J/molk Calculate the AG, at 500 °c with all gases at standard pressure and equilibrium constant K at 500 °C. AG, at 500 °c equals O kJ/mol K at 500 equals tance (kJ/mol) (J/mol-K) O2(g) 205.2 SO2(g) -296.8 248.2 SO3(9) -395.7 240.0 AHO so
Calculate ΔHf°(SO3) from the following information. S(s) + O2(g) → SO2(g) ΔHrxn° = -296.8 kJ SO2(g) + 1/2 O2(g) → SO3(g) ΔHrxn° = -98.9 kJ