Consider the processes A-D below to answer questions a-b for each of it. a) In t

Question

Consider the processes A-D below to answer questions a-b for each of it. a) In the square provided, predict the signs of Delta H degree, Delta S degree and Delta C degree 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. O_2(g) rightarrow 2O(g) B. 2NO(g) + O_2(g) rightarrow 2NO_2(g) (exothermic) C. combustion of propane gas (C_3H_8) to form H_2O(g) and CO_2(g) D. H_2O_2(I) rightarrow H_2O(I) + O_2(g) (exothermic) b) On the graph provided draw the plot free energy (G degree) vs. reaction composition for the reaction D. Label Delta G degree and the position of the equilibrium.

Explanation / Answer

delta G = -ve ; reaction is spontaneous
delta G = +ve ; reaction is non-spontaneous
delta G = 0 ; reaction is at equilibrium

a) O2 (g) ---------> 2O (g)
Energy must be provided or absorbed by the system in order to break the bond , thus delta H = +ve
As the number of atoms after the reaction or transformation increases( delta n gaseous = moles of gaseous products - moles gaseous reactants; = 2-1 =1) ,the disorderness increases, thus
delta S =+ve
As delta G = delta H- Tdelta S
As delta H and delta S both are +ve quantities therefore the sign of delta G and the spontaneity of this reaction depends on Temperature. Higher the value off "T" , more negative delta G becomes. So, the reaction is spontaneous at high temperatures

b) 2NO (g) + O2 (g) ----------> 2 NO2 (g) (exothermic)

As the reaction is exothermic , that means heat is released by the system, thus delta H = -ve
The number of molecules after the reaction decreases, the disorderness decreases( delta n gaseous = moles of gaseous products - moles gaseous reactants; =2-3= -1), thus delta S= -ve
According to the relation between G, H and S stated above, delta G is negative and the system is spontaneous at low temperature.

c) Combustion of propane gas (C3H8 ) to form CO2 and H2O
C3H8 + 5 O2 ---------> 3 CO2 + 4 H2O
As heat is released by the system in this reaction, thus its an exothermic reaction and delta H = -ve
As the number of gaseous products is more than the number of gaseous reactants , thus delta S= +ve
delta G = -ve as delta H is-ve and delta S is +ve, so according to the realtion stated in part (a), at any Temperature the reaction will be spontaneous as delta G is -ve

d) H2O2 (l) -------------> H2O (l) + O2 (g) (exothermic )

As the reaction is exothermic , that means heat is released by the system, thus delta H = -ve
The number of gaseous products is more than the number of gaseous reactants , thus delta S = +ve
delta G = -ve at all temperature as the same reason stated in the last part

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