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mote1985 [20]
3 years ago
7

At what temperature is the following reaction at equilibrium? 2Na, O, + 2H,0 + 4NaOH + O, AH° = -109 kJ, ASO = -133.2 J/K

Chemistry
1 answer:
natali 33 [55]3 years ago
4 0

Answer:

Temperature = 818 K

Explanation:

The given reaction is:

2Na2O2 + H2O \rightarrow 4NaOH + O2

The change in enthalpy is: ΔH°= -109 kJ

The change in entropy is : ΔS° = -133.2 J/k = -0.1332 kJ/K

Based on the Gibbs-Helmholtz equation the standard free energy change is given as:

\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}

At equilibrium ΔG° = 0. therefore:

\Delta H^{0} = T\Delta S^{0}

T=\frac{\Delta H^{0}}{\Delta S^{0}}=\frac{-109}{-0.1332}=818 K

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How many moles of AIC3 are there in 1,119.972 g ?
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3 years ago
1. Show that heat flows spontaneously from high temperature to low temperature in any isolated system (hint: use entropy change
Inga [223]

Answer:

1 ) Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

                 = Q/tc - Q/th

2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

Explanation:

<u>1) To show that heat flows spontaneously from high temperature to low temperature </u>

example :

Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )

Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment

Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )

Δs ( entropy change for hot block ) = - Q / th  ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

                 = Q/tc - Q/th

<u>2) Entropy change for Decomposition of mercuric oxide </u>

2HgO (s) → 2Hg(l) + O₂ (g)

Δs = positive

there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C

hence ΔSdecomposition = S⁻ Hg  -  S⁻ HgO =

Δh of reaction = 181.6 KJ

Temp = 500 + 273 = 773 k

hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

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In the decomposition reaction, 1 mole of water (mw = 18.015 g/mol) was produced for every mole of cuo (mw = 79.545 g/mol) produc
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Reactives -> Products

CuO and water are products.

I found this reaction which has CuO and water as products: decomposition of Cu(OH)2.

Cu(OH)2 -> CuO + H2O

Stoichiometry calculus involve the mole proportions you can see in the reaction: When 1 mole of Cu(OH)2 reacts, 1 mole of CuO and 1 mole of H2O are formed.

Considering the molar masses:

Cu(OH)2 = 83.56 g/mol

CuO = 79.545 g/mol

H2O = 18.015 g/mol

Then: When 83.56 g of Cu(OH)2 react, 79.545 g of CuO and 18.015 g H2O are formed.

You should use that numbers in the rule of three:

79.545 g CuO __________18.015 g water

3.327 g CuO__________ x =3.327*18.015 /79.545 g water 

x= 0.7535 g water




3 0
3 years ago
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