12. Calculate the maximum energy, in the form of labor, involved in the combusti
ID: 1031071 • Letter: 1
Question
12. Calculate the maximum energy, in the form of labor, involved in the combustion of 1.00 mol of C2H6 ethane (g) at 25 ° C and 101.325 kPa (standard conditions).According to your results, will this work be done by the system on the environment or vice versa?
On the other hand, would this process be spontaneous at all temperatures? justify
C2H6(g) + 7/2 O2(g)---->2CO2(g) + 3 H2O(I) delta H•comb= -1560,7 kJ/mol
S•( C2H6(g))= 229,2 J K-1 mol-1 S•(O2(g))=205,2 J K-1 mol-1 S•( CO2(g)) = 213,8 J K-1 mol-1 S•( H2O(l)) = 70,0 J K-1 mol -1 12. Calculate the maximum energy, in the form of labor, involved in the combustion of 1.00 mol of C2H6 ethane (g) at 25 ° C and 101.325 kPa (standard conditions).
According to your results, will this work be done by the system on the environment or vice versa?
On the other hand, would this process be spontaneous at all temperatures? justify
C2H6(g) + 7/2 O2(g)---->2CO2(g) + 3 H2O(I) delta H•comb= -1560,7 kJ/mol
S•( C2H6(g))= 229,2 J K-1 mol-1 S•(O2(g))=205,2 J K-1 mol-1 S•( CO2(g)) = 213,8 J K-1 mol-1 S•( H2O(l)) = 70,0 J K-1 mol -1
According to your results, will this work be done by the system on the environment or vice versa?
According to your results, will this work be done by the system on the environment or vice versa?
On the other hand, would this process be spontaneous at all temperatures? justify
C2H6(g) + 7/2 O2(g)---->2CO2(g) + 3 H2O(I) delta H•comb= -1560,7 kJ/mol
S•( C2H6(g))= 229,2 J K-1 mol-1 S•(O2(g))=205,2 J K-1 mol-1 S•( CO2(g)) = 213,8 J K-1 mol-1 S•( H2O(l)) = 70,0 J K-1 mol -1
Explanation / Answer
C2H6(g) + 7/2 O2(g)---->2CO2(g) + 3 H2O(I) delta H•comb= -1560,7 kJ/mol
delta S = Sproduct - Sreactant
delta S = { 2* S•( CO2(g)) + 3* S•( H2O(l)) } - { S•( C2H6(g)) + 3.5* S•(O2(g)) }
delta S = {2* 213,8 + 3* 70,0 } - { 229,2 +3.5* 205,2 }
delta S = -309.8 J K-1 mol -1
delta G = delta H•comb - T* delta S
delta G = -1560,7*1000 - (25+273.15)*(-309.8) = -468333.13 KJ (This is the enrgy in labor form)
Since our delta G is negetive so the process is spontaneous.
NOTE: I HAVE USED COMMA " , " AS THE POINTS " . " . SO IF THE COMMA IS NOT WRITTEN IN PLACE OF POINTS THEN FOLLOW THE SAME PROCEDURE . YOU WILL GET THE ANSWER ACCORDING TO NEW INPUTS, METHOD WILL REMAIN SAME.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.