Xe(g) + 3F_2 (g) left and right arrow XF_6 (g) Under standard conditions, the en
ID: 931726 • Letter: X
Question
Xe(g) + 3F_2 (g) left and right arrow XF_6 (g) Under standard conditions, the enthalpy change for the reaction going from left to right (forward reaction) is triangle H degree = 294 k J. Is the value of triangle Sdegree, for the above reaction, positive or negative? Justify your answer. The above recation is spontaneous under standard conditions. Predict what will happen to triangle G for this recation as the temperature is increased . Justify your prediction . show how the tempearture at which the reaction changes from spontaneous non-spontaneous can be predicted. What additional information is needed to determine the exact temperature?Explanation / Answer
Given reaction Xe (g) + 3 F2 (g) -----------> XeF6 (g) Ho = -294 kJ
a) So value is negative.
Because 4 moles of gas is converted to 1 mol of gas. Hence entropy decreases.
Therefore, So value is negative.
b) Go = Ho - T So
Given that Ho = -294 kJ and So = -ve
Go = - 294 kJ - T (-ve)
= - 294 kJ + T
If temperature is increased, Go becomes +ve.
Hence, the reaction becomes non-spontaneous.
( Go = - ve , spontaneous
Go = + ve , non- spontaneous )
c)
Go > 0 ( condition for non-spontaneity)
Ho - T So > 0
Ho > T So
T So < Ho
T < Ho/ So
T < (- 294 kJ) / ( So )
So , to determine the exact temperature at which reaction becomes non-spontaneous So is required.
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