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AP Chemistry
Calculate the delta H rxn for the following reaction: CH4(g)+4Cl2(g)-->CCl4(g)+4HCl Use the following reactions and given delta H's: 1) C(s)+2H2(g)-->CH4(g) delta H= -74.6 kJ 2) C(s)+2Cl2(g)-->CCl4(g) delta H= -95.7 kJ 3)
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Which of the following reactions are spontaneous (favorable)? A] 2Mg(s) + O2(g) -> 2MgO(s) Delta G = -1137 kJ/mol .. B] NH3(g) + HCl(g) -> NH4Cl(s) Delta G = -91.1 kJ/mol ... C] AgCl(s) -> Ag+(aq) + Cl-(aq) Delta G = 55.6 kJ/mol
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For the reaction described by the chemical equation: 3C2H2(g) -> C6H6(l) .. Delta H rxn = -633.1 kJ/mol a) Calculate the value of Delta S rxn at 25.0 C... b) Calculate Delta G rxn... c) In which direction is the reaction, as
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Calculate the enthalpy change for the reaction: 2S(s) + 3O2(g) --> 2SO3(g) Use the following thermochemical equations: Reaction S(s) + O2(g) --> SO2(g) Delta H(in kJ) -296.8 2SO2(g) + O2(g) --> 2SO3(g) Delta H(in kJ) -197.0
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Given the thermodynamic data in the table below, calculate the equilibrium constant for the reaction: 2SO2(g)+O2--> 2SO3 Substance (DeltaH^o) (Delat S^o) SO2 -297 249 O2 0 205 SO3 -395 256 Answer (it was given) 2.32x10^24 Even
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How would you calculate delta H for the following reaction: 2P + 5Cl2 -> 2PCl5 Given: PCl3 + Cl2 -> PCl5 Delta H=-87.9 kJ 2P + 3Cl2 -> 2PCl3 Delta H=-574 kJ I have no clue how to do this since I was absent. Can someone help?
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Reaction 1: NaOH + HCL --> H20 + NaCl delta H of -100.332 kj/mol Reaction 2: NaOH + NH4Cl --> NH3 + H20 : delta H of 358.639 kj/mol Reaction 3: HCl + NH3 --> NH4Cl : delta H of -51.701 kj/mol Use your answers from question 2 above
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Calculate the delta H and delta S for the reaction. From there, calculate the delta G at 25 degrees Celsius. Predict whether it the reaction is spontaneous or non-spontaneous under standard conditions. CH3OH(l)+ O2(g)--> HCO2H(l)+
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Please help me N2O5(g) + H2O(l) --> 2HNO3(l) delta H° = -76.2 kJ H2O(l) --> H2(g) + 1/2O2(g) delta H° = 286.0 kJ 1/2N2(g) + 3/2O2(g) + 1/2H2(g) --> HNO3(l) delta H° = -174.0 kJ Calculate delta H° for the reaction 2N2O5(g) -->
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Given overall reaction: P4 + 10Cl2 ---> 4PCl5 Delta H for the reaction = ? PCl5 -----> PCl3 + Cl2 Delta H = +157 kJ P4 + 6Cl2 ----> 4PCl3 Delta H = -1207 kJ Calculate the Delta H for the overall reaction.
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At 25 C the following heats of reaction are known: 2CIf (g)+ O2(g) -> CL2O (g) + f2O Delta H rxn= 167.4 kj/mol 2CIf3(g) + 2O2 (g)-> Cl2O(g) + 3F2O(g) Delta H rxn= 341.4 2F2(g) + O2(g) -> 2F2O(g) Delta H rxn= -43.4kj/mol at the
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