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Damm [24]
3 years ago
7

During an endothermic reaction, H for the reactants was −600 kJ/mol. Which of the following statements is correct about the H fo

r the products and the comparison of the energy in bonds?
A. It is less than −600 kJ/mol, and the amount of energy required to break bonds is greater than the amount of energy released in forming bonds.

B. It is less than −600 kJ/mol, and the amount of energy required to break bonds is less than the amount of energy released in forming bonds.

C. It is greater than −600 kJ/mol, and the amount of energy required to break bonds is greater than the amount of energy released in forming bonds.

D. It is greater than −600 kJ/mol, and the amount of energy required to break bonds is less than the amount of energy released in forming bonds.
Chemistry
2 answers:
o-na [289]3 years ago
7 0

Answer:

C. It is greater than -600 kJ/mol, and the amount required to break bonds is greater than the amount of energy released in forming bonds.  

Explanation:

As a reaction occurs, bonds of reactants are broken and new bonds are formed in products. During bond breaking energy is absorbed while during bond formation energy is released. If the energy released during bond formation is greater than the energy absorbed during bond breaking, then the reaction is exothermic. If the energy released during bond formation is less than energy absorbed during bond breaking, then the reaction is endothermic.

If reactants have less energy than products, it means more energy is absorbed than released thus it is an endothermic reaction.

If reactants have more energy than products, it means more energy is released than absorbed thus it is an exothermic reaction.

anastassius [24]3 years ago
7 0

Answer:

its c

Explanation:

i took the exam

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VikaD [51]

Answer : The value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

Explanation :

The balanced cell reaction will be,

Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)

The half-cell reactions are:

Oxidation reaction (anode) : Pb(s)\rightarrow Pb^{2+}(aq)+2e^-

Reduction reaction (cathode) : 2Ag^+(aq)+2e^-\rightarrow 2Ag(g)

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

where,

\Delta G^o = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential of the cell = 0.93 V

Now put all the given values in the above formula, we get:

\Delta G^o=-2\times 96500\times 0.93

\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol

Now we have to calculate the value of 'K'.

\Delta G^o=-RT\ln K

where,

\Delta G_^o =  standard Gibbs free energy  = -180 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K

K=3.6\times 10^{31}

Therefore, the value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

5 0
3 years ago
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