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DerKrebs [107]
3 years ago
10

Which bond from each set has the greatest bond energy and why?. A.) N-O or N-F. B.) B-F or B-Cl. C.) B-C or O-F. D.) C-O or C=O.

E.) P-Br or P-Cl
Chemistry
2 answers:
Vikki [24]3 years ago
7 0

Answer:

    A) N-F,  B) B-F,  C) O-F, D) C=O, E) P-Cl.

Explanation:

Bond energy is the energy that is to break the molecule apart. Bond energy affected by the electronegativity and no. of bonds.

A) N-F bond is more powerful than N-O because Fluorine is the highest electronegative element in the periodic table.

B) B-F has strongest bond than B-Cl, due to higher electronegativity of fluorine.

C) O-F has highest bond energy than B-C, because O is strong covalently bonded with fluorine which is the highest electronegative.

D) C=O has highest bond energy, because double bond has required more energy to break than the single bond.

E) P-Cl has high bond energy than the P-Br, because Cl has small size and more electronegative than Br.

Solnce55 [7]3 years ago
5 0
A) N-F becasue F ismore electronegative so the covalent bond is stronger
B) B-Cl because Cl is a smaller atom than Br so the covalent bond is stronger 
C) O-F because the atoms are more electronegative so the bond is stronger.
D) C=O because the double bond is stronger than the single bond.
E) P-Cl because Cl is smaller than Br so the electrostatic force of attraction between the positive nucleus and the bonding electrons is stronger, thus the covalent bond is stronger 

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CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7
padilas [110]

<u>Answer:</u> The \Delta G for the reaction is 54.425 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}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm

Putting values in above equation, we get:

K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85

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 = 20.85

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol

Hence, the \Delta G for the reaction is 54.425 kJ/mol

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