Chemistry, 26.10.2019 05:43, lae270

Above what temperature does the following reaction become nonspontaneous? feo(s) + co(g) → co2(g) + fe(s) δh = -11.0 kj; δs = -17.4 j/k 191 k 298 k

Answers: 2

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Answers

answered: gonzalesalexiaouv1bg

632 K

Explanation:

1) Reaction (given):

FeO(s) + CO(g) → CO₂(g) + Fe(s) ΔH = -11.0 kJ;

ΔS = -17.4 J/K

2) Spontaneous reactions (ΔG < 0)

The sign of the change in free energy (ΔG) tells if a reaction is spontaneous or not at a given temperature and constant pressure.

If ΔG is negative ( ΔG < 0) the reaction is spontaneous; if ΔG is positive ( ΔG > 0) the reaction is nonspontaneous.

Then, determine the temperature at which ΔG becomes zero, which is the temperature at which the reaction becomes nonspontaneous.

3) ΔG⁰ = ΔH⁰ - TΔS⁰

ΔG⁰ = -11,000 J - T (-17.4 J/ K) [ -11.0 kJ were converted to J]

0 = - 11,000 J + 17.4 T J/K

11,000 J = 17.4 T J/ k

T = 11,000 kJ / (17.4 J/K) = 632 K ← answer

answered: harvzoie

The following reaction become nonspontaneous at 632.18 K

Further explanation

Gibbs free energy is the maximum possible work given by chemical reactions at constant pressure and temperature. 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= standart)

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)

Reaction :

FeO(s) + CO(g) → CO2(g) + Fe(s)

ΔH = -11.0 kJ= -11000 J; ΔS = -17.4 J/K

ΔG =-11000 -(T.-17.4 J/K)

0 = -11000 + T17.4

T17.4 = 11000

T =632.18

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 of formation,nonspontaneous

answered: Guest

That result from subduction of oceanic plates beneath lighter continental plates. most of the earth's volcanoes are located around the pacific ring of fire because that the location of most of the earth's subduction zones.

answered: Guest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.