Tungsten is usually produced by the reduction of WO3 with the hydrogen as shown

Question

Tungsten is usually produced by the reduction of WO3 with the hydrogen as shown by the equation:
WO3 (g)+ 3H2 (g)----W (s) + 3 H20 (g)
The enthalpy for the reaction, delta H rxn is +114.5 kJ. The entropy for the reaction, ?DeltaS
is +132J/K.

Myquestions are:
USe the enthalpy and entropy values given to calculate the delta G for this reaction at 25 degrees C.

First: Show the relationship among the three thermodynamic functions before doing the calculation.
Then: Explain whether or not this process is spontaneous at 25 degrees C.
Next: Explain what can be done to make the process spontaneous
Last: Give the name for delta G

Explanation / Answer

If delta G (free energy) is positive (> 0), the reaction is endothermic and non-spontaneous. If delta G (free energy) is negative (< 0), the reaction is exothermic and spontaneous. If delta H (enthalpy change) is positive (> 0), the reaction absorbs heat and is endothermic. If delta H (enthalpy change) is negative (< 0), the reaction releases heat and is exothermic. If delta S (entropy) is positive (> 0), there will be an increase in disorder and the reaction will be spontaneous if the enthalpy change is negative. If the enthalpy change is positve, the reaction will be spontaneous only if the temperature is high enough. If delta S (entropy) is negative (< 0), there will be a decrease in disorder and the reaction will be non-spontaneous. If the enthalpy change is negative, the reaction will be spontaneous only if the temperature is high enough. If the enthalpy change is positive, the reaction will be non-spontaneous. A non-spontaeous reaction is one that needs a constant supply of free energy to keep it running. delta H = 114.5 kJ = 114,500 J delta S = 132 J/K delta G = delta H - T * delta S delta G = (114,500 J) - (298.15 K)(132 J/K) delta G = 75,144.2 J = 75.14 kJ --> the reaction is non-spontaneous because delta G is > 0. The temperature will tell us when the reaction is spontaneous. To solve for the temperature at which the reaction will be spontaneous, we sub in 0 for delta G: delta G = (114,500 J) - T(132 J/K) 114,500 J = - T(132 J/K) T = 867.42 K --> the reaction will be spontaneous above 867.42 K. When you sub this temperature in, you get 0.56. If you plug in values higher than this temperature, they will eventually be < 0. delta G is Gibbs free energy. Hope this helps! :-)

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