Question

Type the correct answer in the box. Express the answer to two significant figures.

Answer 1

Delta enthalpy = 2x386-3x1x432-3x942=-3350kJ/mol

Answer 2

Answer:

The enthalpy change for the reaction is -78 kJ

Explanation:

In the given reaction, 1 mol of $N\equiv N$ and 3 moles of H-H bonds are broken. Also 6 moles of N-H bonds are formed as 1 mol of $NH_{3}$ contains 3 moles of N-H bonds.

Now, energy is released during bond formation. So, for this case, sign of bond energy will be negative. Also energy if supplied during bond breaking. So, for this case, sign of bond energy will be positive.

Hence change in enthalpy of reaction= $(1\times bond enrgy of N\equiv N)+(3\times bond energy of H-H)-(6\times bond energy of N-H)$=$(1\times 942)+(3\times 432)-(6\times 386)kJ= -78 kJ$