Question

Chemistry help !! 10 points:)? Without doing any calculations, determine the sign of ΔSsys and ΔSsurr for each of the chemical reactions below. 1) 2CO(g)+O2(g)⇌2CO2(g)ΔH∘rxn= -566.0 kJ -566.0 a) ΔSsys>0, ΔSsurr>0 b) ΔSsys<0, ΔSsurr>0 c) ΔSsys>0, ΔSsurr<0 d) ΔSsys<0, ΔSsurr<0 2) 2NO2(g)→2NO(g)+O2(g)ΔH∘rxn= +113.1 kJ +113.1 a) ΔSsys>0, ΔSsurr>0 b) ΔSsys<0, ΔSsurr>0 c) ΔSsys>0, ΔSsurr<0 d) ΔSsys<0, ΔSsurr<0 3) 2H2(g)+O2(g)→2H2O(g)ΔH∘rxn= -483.6 kJ -483.6 a) ΔSsys>0, ΔSsurr>0 b) ΔSsys<0, ΔSsurr>0 c) ΔSsys>0, ΔSsurr<0 d) ΔSsys<0, ΔSsurr<0 4) CO2(g)→C(s)+O2(g)ΔH∘rxn= +393.5 kJ +393.5 a) ΔSsys>0, ΔSsurr>0 b) ΔSsys<0, ΔSsurr>0 c) ΔSsys>0, ΔSsurr<0 d) ΔSsys<0, ΔSsurr<0 In addition, predict under what temperatures (all temperatures, low temperatures, or high temperatures), if any, the reaction in part A, B, C and D will be spontaneous. a) The reaction is spontaneous at all temperatures. b) The reaction is spontaneous at low temperatures. c) The reaction is spontaneous at high temperatures. d) The reaction is nonspontaneous at all temperatures.

Answer 1

I think the equation for finding ΔS(sur) is ΔS(surr)= -ΔH/T so if the reaction is exothermic the ΔS(surr) should be positive and if the reaction is endothermic the ΔS(surr) should be negative.  The sign for ΔS(sys) can be found by seeing if the products side or reactants side has more moles of gas (or ions if being dissolved).  If the products side has more moles of gas  (or ions if being dissolved) than the reactants side, the ΔS(sys) is positive.  If the reactants side has more moles of gas (or ions if being dissolved) than the products the ΔS(sys) is negative.

To determine what temperatures the reaction is spontaneous you need to use the equation ΔG(sys)=ΔH(sys)-TΔS(sys).  The reaction is spontaneous if the ΔG is negative so if you have a negative ΔH and negative ΔS the reaction is only spontaneous at low temperatures.  If you have a positive ΔH and negative ΔS the reaction is never spontaneous.  If you have a negative ΔH and positive ΔS the reaction is always spontaneous.  if you have a positive  ΔH and positive ΔS the reaction is spontaneous at high temperatures.

To do the first reaction.  You have 3 moles of gas going to 2 moles of gas so the ΔS(sys) is negative since it is going from a state of higher disorder to lower disorder.  Since the reaction is exothermic the ΔS(surr) is positive.  Since ΔH is negative and ΔS is negative the reaction is spontaneous at only low temperatures ( has a negative ΔG).

Answer 2

Part A : ΔSsys<0, ΔSsurr>0

Part B : ΔSsys> 0, ΔSsurr <0

Part C : ΔSsys <0, ΔSsurr> 0

Part D : ΔSsys> 0, ΔSsurr <0

Further explanation

Gibbs free energy can be used to determine the spontaneity of a reaction

If the Gibbs free energy value is <0 (negative) then the chemical reaction occurs spontaneously. If the change in free energy is zero, then the chemical reaction is at equilibrium, if it is> 0, the process is not spontaneous

Free energy of reaction (G) is the sum of its enthalpy (H) plus the product of the temperature and the entropy (S) of the system

Can be formulated: (at any temperature)

$\large {\boxed {\bold {\Delta G = \Delta H-T. \Delta S}}}$

or at (25 Celsius / 298 K, 1 atm = standard)

ΔG ° reaction = ΔG ° f (products) - ΔG ° f (reactants)

Under standard conditions:

∆G ° = ∆H ° - T∆S °

The value of °H ° can be calculated from the change in enthalpy of standard formation:

∆H ° (reaction) = ∑H ° (product) - ∑ H ° (reagent)

The value of ΔS ° can be calculated from standard entropy data

∆S ° (reaction) = ∑S ° (product) - ∑ S ° (reagent)

or can be formulated:

ΔS = ΔQ / T

• Part A

2CO (g) + O2 (g) ⇌2CO2 (g) ΔH∘rxn = -566.0 kJ

ΔH = -, exothermic, heat moves from the system to the environment so that ΔSsys <0, ΔSsurr> 0

• Part B

2NO2 (g) → 2NO (g) + O2 (g) ΔH∘rxn = +113.1 kJ

ΔH = +, endothermic, heat moves from the environment to the system so ΔSsys> 0, ΔSsurr <0

• Part C

2H2 (g) + O2 (g) → 2H2O (g) ΔH∘rxn = -483.6 kJ

ΔH = -, exothermic, heat moves from the system to the environment so that ΔSsys <0, ΔSsurr> 0

• Part D

CO2 (g) → C (s) + O2 (g) ΔH∘rxn = +393.5 kJ

ΔH = +, endothermic, heat moves from the environment to the system so ΔSsys> 0, ΔSsurr <0

From the Gibbs equation can be concluded:

an endothermic reaction is spontaneous at high temperature and exothermic reaction is spontaneous at low temperature

The reaction part A (exothermic) is spontaneous at low

temperatures.

The reaction part B (endothermic) is spontaneous at high

temperatures.

The reaction part C (exothermic) is spontaneous at low

temperatures.

The reaction part D (endothermic) is spontaneous at high

temperatures.

Learn more

Delta H solution

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an exothermic reaction

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 as endothermic or exothermic

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an exothermic dissolving process

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Keywords: the standard gibbs free energy