For the reaction 2BrF3 (g) → Br2(g) + 3F₂ (g) AG = 451 kJ and AH° 542 kJ at 338 K and 1 atm. This reaction is favored under standard conditions at 338 K. The entropy change for the reaction of 1.54 moles of BrF3 (9) at this temperature would be J/K.

Spontaneous When entropy is positive and enthalpy is negative, a process is spontaneous. This means the reaction proceeds on its own under standard conditions, giving off energy as the reaction proceeds. Spontaneous at High Temperatures When entropy is positive and enthalpy is positive, a process is spontaneous at high temperatures because the products in the system are able to use thermal energy from the surroundings in order to react. Thermal energy increases the kinetic energy of the molecules. An example of this is ice melting. Ice will not melt if the temperature is not high enough. Spontaneous at Low Temperatures When entropy is negative and enthalpy is negative, a process is spontaneous at low temperatures. This type of reaction will release energy into the surroundings, and the ability to release this heat is important. If the temperature of the surrounding system is too high, the reactants have to overcome this energy in order to react and release their own heat energy. An…

For the reaction N₂ (g) + 2O2(g) → 2NO2(g) - 122 J/K AH° = 66.4 kJ and AS = The equilibrium constant for this reaction at 343.0 K is Assume that AH° and AS are independent of temperature.

A student determines the value of the equilibrium constant to be 6.29×104 for the following reaction.S(s,rhombic) + 2CO(g)SO2(g) + 2C(s,graphite)Based on this value of Keq:G° for this reaction is expected to be (greater, less) fill in the blank  than zero.Calculate the free energy change for the reaction of 1.85 moles of S(s,rhombic) at standard conditions at 298K.    G°rxn = _____ kJ