Question

Part1: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g) ΔH∘rxn=−2044kJ

A)ΔSsys>0

or

B) ΔSsys<0

Part 2:

A)ΔSsurr>0

or

B)ΔSsurr<0

Part 3)

A)The reaction is spontaneous at all temperatures.

B)The reaction is spontaneous at high temperatures.

C)The reaction is spontaneous at low temperatures.

D)The reaction is nonspontaneous at all temperatures.

Part 4) N2(g)+O2(g)→2NO(g) ΔH∘rxn=+182.6kJ

A)ΔSsys>0

or

B) ΔSsys<0

Part 5)

A)ΔSsurr>0

or

B)ΔSsurr<0

Part 6)

A)The reaction is spontaneous at all temperatures.

B)The reaction is spontaneous at high temperatures.

C)The reaction is spontaneous at low temperatures.

D)The reaction is nonspontaneous at all temperatures. Part 7) 4NH3(g)+5O2(g)→4NO(g)+6H2O(g) ΔH∘rxn=−906kJ A)ΔSsys>0 or B) ΔSsys<0 Part 8) A)ΔSsurr>0 or B)ΔSsurr<0 Part 9) A)The reaction is spontaneous at all temperatures. B)The reaction is spontaneous at high temperatures. C)The reaction is spontaneous at low temperatures. D)The reaction is nonspontaneous at all temperatures. Part 10) 2N2(g)+O2(g)→2N2O(g) ΔH∘rxn=+163.2kJ A)ΔSsys>0 or B) ΔSsys<0 Part 11) A)ΔSsurr>0 or B)ΔSsurr<0 Part 12) A)The reaction is spontaneous at all temperatures. B)The reaction is spontaneous at high temperatures. C)The reaction is spontaneous at low temperatures. D)The reaction is nonspontaneous at all temperatures.

Answer 1

Part 1)

C3H8(g)+5O2(g)→3CO2(g)+4H2O(g) ΔH∘rxn=−2044kJ

So more number of gaseous molecules on product side so the entropy change of system will be positive (S >0)

Delta S system> 0

Part 2) Delta S surr > 0 [exothermic reaction]

Part 3)

Delta G = Delta H - T Delta S

Delta H = negative

reaction will be spontaneous if Delta G < 0

Which is possible at all temperatures

Part 7) 4NH3(g)+5O2(g)→4NO(g)+6H2O(g) ΔH∘rxn=−906kJ

The number of gaseous molecules in the product side is more so entropy change of system will be > 0

PArt 8) Delta S > 0

Part 9)

Delta G = Delta H - T Delta S

Delta H = negative

reaction will be spontaneous if Delta G < 0

Which is possible at all temperatures

Part 10

2N2(g)+O2(g)→2N2O(g) ΔH∘rxn=+163.2kJ

Delta S system < 0 as there are less number of gaseous molecules on the product side

Part 11)

Delta S surrounding < 0

Part 12)

Delta G = Delta H - T Delta S

Delta H = Positive

reaction will be spontaneous if Delta G < 0

Which is possible at high temperatures