.127. Use average bond energies to estimate the enthalpy changes of the following reactions: N2(g)+3 H2(g)2...

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.127. Use average bond energies to estimate the enthalpy changes of the following reactions: N2(g)+3 H2(g)2 NH(g) b. N2(g)+2

9.103. Consider the following molecular ions: N2, O2, C2, and Br22. Using MO theory, (a) write their orbital electron configu

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Answer 1

Answer #1

#127: (a): N2 has 1 NEN , H2 has 1 H-H and NH3 has 3 N - H single bonds.

Hence 2NH3 has a total of 2*3 = 6 N-H single bonds

$\Delta$ _rH = $\sum$ (Bond energy of all reactants) - $\sum$ (Bond energy of all products)

=> $\Delta$ _rH = [1 mol * (BE, N2) + 3 mol*(BE, H2)] - [ 6 mol * (BE, N-H)]

=> $\Delta$ _rH = [1mol * 945 kJ/mol + 3 mol*436 kJ/mol] - [6 mol * 450.2 kJ/mol]

=> $\Delta$ _rH = 945 kJ + 1308 kJ - 2701.2 kJ

=> $\Delta$ _rH = - 448.2 kJ (Answer)

(b):

N2 has 1 NEN , H2 has 1 H-H. H2NNH2 has 4 N-H single and 1 N-N single bond

$\Delta$ _rH = $\sum$ (Bond energy of all reactants) - $\sum$ (Bond energy of all products)

=> $\Delta$ _rH = [1 mol * (BE, N2) + 2 mol*(BE, H2)] - [ 1 mol * (BE, N-N) + 4 mol*(BE, N-H)]

=> $\Delta$ _rH = [1mol * 945 kJ/mol + 2 mol*436 kJ/mol] - [1 mol * 271.5 kJ/mol + 4 mol* 450.2 kJ/mol]

=> $\Delta$ _rH = 945 kJ + 872 kJ - 271.5 kJ - 1800.8 kJ

=> $\Delta$ _rH = - 255.3 kJ (Answer)

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Answer 2

Answer #1

#127: (a): N2 has 1 NEN , H2 has 1 H-H and NH3 has 3 N - H single bonds.

Hence 2NH3 has a total of 2*3 = 6 N-H single bonds

$\Delta$ _rH = $\sum$ (Bond energy of all reactants) - $\sum$ (Bond energy of all products)

=> $\Delta$ _rH = [1 mol * (BE, N2) + 3 mol*(BE, H2)] - [ 6 mol * (BE, N-H)]

=> $\Delta$ _rH = [1mol * 945 kJ/mol + 3 mol*436 kJ/mol] - [6 mol * 450.2 kJ/mol]

=> $\Delta$ _rH = 945 kJ + 1308 kJ - 2701.2 kJ

=> $\Delta$ _rH = - 448.2 kJ (Answer)

(b):

N2 has 1 NEN , H2 has 1 H-H. H2NNH2 has 4 N-H single and 1 N-N single bond

$\Delta$ _rH = $\sum$ (Bond energy of all reactants) - $\sum$ (Bond energy of all products)

=> $\Delta$ _rH = [1 mol * (BE, N2) + 2 mol*(BE, H2)] - [ 1 mol * (BE, N-N) + 4 mol*(BE, N-H)]

=> $\Delta$ _rH = [1mol * 945 kJ/mol + 2 mol*436 kJ/mol] - [1 mol * 271.5 kJ/mol + 4 mol* 450.2 kJ/mol]

=> $\Delta$ _rH = 945 kJ + 872 kJ - 271.5 kJ - 1800.8 kJ

=> $\Delta$ _rH = - 255.3 kJ (Answer)

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Answer 3

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.127. Use average bond energies to estimate the enthalpy changes of the following reactions: N2(g)+3 H2(g)2...

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.127. Use average bond energies to estimate the enthalpy changes of the following reactions: N2(g)+3 H2(g)2...

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