Predict the signs of deltaH delta S and delta G classify each as spontaneous at
ID: 1055075 • Letter: P
Question
Predict the signs of deltaH delta S and delta G classify each as spontaneous at all temperature; never spontaneous...
3. Consider the processes A-D below to answer questions a-b for each of it. (2 pt) a) In the square provided, predict the signs ofAH°, AS° and AG and classify each as: Spontaneous at all temperature; Never spontaneous; Spontaneous at high temperatures; nonspontaneous at low temperatures; OR Spontaneous at low temperatures; nonspontaneous at high temperatures. A. 02(g)20(g) B. 2N0(g) + O2(g) 2NO2(g) (exothermic) C. combustion of propane gas (C3Hs) to form H20(g) and CO2(g) D. H202 (1 - H2O) + O2(g) (exothermic)Explanation / Answer
The relation between free energy and entropy is given by
Delta G = delta H - T.Delta S
A) O2(g) -----> 2O(g)
As we are breaking the bond between two oxygen atoms delta H is positive (endothermic)
delta S is positive (number of gas moles increases)
Thus delta G is positive at low temperatures and negative at high temperatures.
Thus the reaction is non-spontaneous at low temperautres and spontaneous at high temperatures.
B) 2NO(g) + O2(g) --------> 2NO2(g)
exothermic reaction thus delta H is negative (exothermic)
delta S is also negative as the number of gaseous molecules of products less than reactants (3>2)
Thus the Delta G is -ve at low T and +ve at high temperatures.
Thus the reaction is spontaneous at low temperatures and non-spontaneous at high temperatures.
C) C3H8(g) + 5 O2 (g) ---------> 3CO2 (g) + 4H2O(g)
Delta H is -ve (combustion is exothermic)
Delta S is +ve [ gas products (3+4) > reactants(1+5) ]
Thus Delta G is always negative.
Hence the reaction is spontaneous at all temperatures.
D) H2O2(l) ------> H2O(l) + O2(g)
Delta H is negative (exothermic)
Delta S is +ve (gas productsone)
delta G is always negative.
Hence the reaction is spontaneous at all temperatures