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dezoksy [38]
3 years ago
6

Calculate the net change in energy for the following reaction:

Chemistry
1 answer:
AlekseyPX3 years ago
3 0

Answer:

-1078 kJ/mol.

Explanation:

Consider this reaction in three steps:

  1. Break everything down into atoms: 2 Na(g) + 2 H(g) + 2 Cl(g);
  2. Turn 2 Na(g) and 2 Cl(g) into 2 Na⁺(g) and 2 Cl⁻(g);
  3. Combine everything to form 2 NaCl(s) and H₂(g).

Sources of enthalpy changes in the first step:

  • Sublimate 2 Na(s);
  • Break two H-Cl bonds.

That corresponds to an enthalpy change of

\rm 2\times 97 + 2\times 427 = 1,048\;kJ\cdot mol^{-1}.

Sources of enthalpy changes in the second step:

  • Remove one electron for each of the two Na(g);
  • Add one electron to each of the two Cl(g).

That corresponds to an enthalpy change of

\rm 2\times 496 + 2\times (-349) = 294\; kJ\cdot mol^{-1}.

Sources of enthalpy changes in the third step:

  • Bring 2 Na⁺(g) and 2 Cl⁻(g) together to form 2 NaCl(s) (which corresponds to twice the lattice enthalpy of NaCl);
  • Bring 2 H(g) together to form one H₂(g).

That corresponds to an enthalpy change of

\rm 2\times (-778) + 2\times (-432) = -2,420\;kJ\cdot mol^{-1}.

Take the sum of the enthalpy changes of the three steps to find the enthalpy change of the overall reaction:

\rm 1,048 + 294 + (-2,420) = -1078\; kJ\cdot mol^{-1}.

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Answer: Option (d) is the correct answer.

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                                                          = 2.66 - 0.93

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So, we can see that highest electronegativity difference is 1.73 and it is shown by NaI molecule.

Thus, we can conclude that a group 1 alkali metal bonded to iodide, such as NaI has the greatest electronegativity difference between the bonded atoms.

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