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beks73 [17]
3 years ago
15

Consider the following reaction: CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at25 ∘C under these conditions: PCO2PCC

l4PCOCl2===0.140atm0.180atm0.760atm ΔG∘f for CO2(g) is −394.4kJ/mol, ΔG∘f for CCl4(g) is −62.3kJ/mol, and ΔG∘f for COCl2(g) is −204.9kJ/mol. Express the energy change in kilojoules per mole to one decimal place.
Chemistry
1 answer:
Salsk061 [2.6K]3 years ago
7 0

<u>Answer:</u> The \Delta G for the reaction is 54.6 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

  • To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

  • The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.760atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.180atm

Putting values in above equation, we get:

K_p=\frac{(0.760)^2}{0.140\times 0.180}\\\\K_p=22.92

  • To calculate the Gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 22.92

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(22.92))\\\\\Delta G=54659.78J/mol=54.6kJ/mol

Hence, the \Delta G for the reaction is 54.6 kJ/mol

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yanalaym [24]

0.309g

Explanation:

Mass of evaporating dish = 6.251g

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   solving for mass of hydrate before heating;

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 Mass of hydrate before heating = 16.864g - 6.251g = 10.613‬g

Upon heating ;

   Mass of hydrate plus dish = 11.3g

Unknown:

number of moles of water removed from the hydrate = ?

We need to find the mass of water removed from the hydrate;

Mass of hydrate after heating =mass of hydrate plus dish after heating - mass of evaporating dish

mass of hydrate after heating =  11.3g - 6.251g = 5.049g

Mass of water lost = Mass of hydrate before heating - mass of hydrate after heating

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    = 5.564g

Number of moles of water = \frac{mass}{molar mass}  = \frac{5.564}{18}

number of moles of water = 0.309moles

learn more:

Number of moles brainly.com/question/13064292

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5 0
3 years ago
What volume will water vapor form when 40 liters of hydrogen completely reacts with 25 liters of oxygen? Assume the temperature
Brut [27]
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number of moles in reaction                 2 mol                  1 mol       2 mol
number of liters in the reaction         2*22.4 L                1*22.4 L    2*22.4L
We can see that volumes of the gases are proportional to coefficients in the reaction ( if gases are under the same conditions), so we can write

                              2H2           +           O2 ---->2H2O  
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     given                 40 L                      ( 25 L)     40 L

We can see that we have excess of O2,
because if 2 L H2 are needed 1 L O2, then 40 L of H2 are needed 20 L O2.

So, limiting reactant is H2, and we will need to calculate Volume of H2O using H2.
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3 years ago
A mixture of He He , N 2 N2 , and Ar Ar has a pressure of 24.1 24.1 atm at 28.0 28.0 °C. If the partial pressure of He He is 301
Karo-lina-s [1.5K]

Answer:

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Explanation:

<em>A mixture of He, N₂, and Ar has a pressure of 24.1 atm at 28.0 °C. If the partial pressure of He is 3013 torr and that of Ar is 2737 mm Hg, what is the partial pressure of N₂?</em>

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pAr=2737mmHg.\frac{1atm}{760mmHg} =3.60atm

From [1],

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