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gtnhenbr [62]
3 years ago
13

The average bond energy (enthalpy) for a C=C double bond is 614 kJ/mol and that of a C−C single bond is 348 kJ/mol. Estimate the

energy needed to break only the π bond of the double bond of 2-butene. Express your answer numerically in joules per molecule.
Chemistry
1 answer:
AysviL [449]3 years ago
7 0

Answer:

4.42x10⁻¹⁹ J/molecule

Explanation:

At a double bond, there's sigma and a pi bond, and at a single bond, there's only a sigma bond. Thus, if the energy to break both sigma and pi is 614 kJ/mol, and the energy to break only the sigma bond is 348 kJ/mol, the energy to break only the pi bond is:

E = 614 - 348 = 266 kJ/mol

Knowing that 1 kJ = 1000 J, E = 266,000 J/mol

By Avogadro's number, 1 mol = 6.02x10²³ molecules, thus:

E = 266,000 J/mol * 1mol/6.02x10²³ molecules

E = 4.42x10⁻¹⁹ J/molecule

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<u>Answer:</u> The correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

<u>Explanation:</u>

We are given:

Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V

Reduction half reaction: Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V       ( × 3)

Net equation:  3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.80-(-0.73)=1.53V

Hence, the correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

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