Chemistry 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.

Part A : ΔSsys<0, ΔSsurr>0

Part B : ΔSsys> 0, ΔSsurr <0

Part C : ΔSsys <0, ΔSsurr> 0

Part D : ΔSsys> 0, ΔSsurr <0

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)

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

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

Under standard conditions:

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:

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

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

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

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.

Delta H solution

link

an exothermic reaction

link

 as endothermic or exothermic

link

an exothermic dissolving process

link

Keywords: the standard gibbs free energy

8 m/s because for him to get the same amount of energy as ray is putting out, bobby, who weighs half as much as ray, needs to run twice as fast.