Type the correct answer in the box. Express the answer to two significant figures Given: [ mathrm(N)_(2)+3 mathrm(H)_(2) arrow 2 mathrm(NH)_(3) ] Bond & }(c) Bond Energy (mathrm(kJ) / mathrm(mol)) mathrm(N)=mathrm(N) & 942 mathrm(H)-mathrm(H) & 432 mathrm(N)-mathrm(H) & 386 Use the bond energies to calculate the change in enthalpy for the reaction. The enthalpy change for the reaction is kilojoules.

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Type the correct answer in the box. Express the answer to two significant figures Given: [ mathrm(N)_(2)+3 mathrm(H)_(2) arrow 2 mathrm(NH)_(3) ] Bond & }(c) Bond Energy (mathrm(kJ) / mathrm(mol)) mathrm(N)=mathrm(N) & 942 mathrm(H)-mathrm(H) & 432 mathrm(N)-mathrm(H) & 386 Use the bond energies to calculate the change in enthalpy for the reaction. The enthalpy change for the reaction is kilojoules.

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To calculate the change in enthalpy (\(\Delta H\)) for the given reaction, we will follow these steps:1. Calculate the total energy required to break the bonds in the reactants.2. Calculate the total energy released when the bonds in the products are formed.3. Use the relationship \(\Delta H = \text{energy required to break bonds} - \text{energy released when new bonds form}\).Given reaction:\[\mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3}\]From the table, we have the following bond energies:- \(\mathrm{N}=\mathrm{N}\): 942 kJ/mol- \(\mathrm{H}-\mathrm{H}\): 432 kJ/mol- \(\mathrm{N}-\mathrm{H}\): 386 kJ/molNow, let's calculate the energy required to break the bonds in the reactants:- For \(\mathrm{N}_{2}\), we have one \(\mathrm{N}=\mathrm{N}\) bond, so the energy required is \(1 \times 942\) kJ/mol.- For \(3 \mathrm{H}_{2}\), we have three \(\mathrm{H}-\mathrm{H}\) bonds, so the energy required is \(3 \times 432\) kJ/mol.Total energy required to break the bonds in the reactants:\[1 \times 942 + 3 \times 432 = 942 + 1296 = 2238 \text{ kJ/mol}\]Next, let's calculate the energy released when the bonds in the products are formed:- For \(2 \mathrm{NH}_{3}\), we have six \(\mathrm{N}-\mathrm{H}\) bonds (each \(\mathrm{NH}_{3}\) has three \(\mathrm{N}-\mathrm{H}\) bonds), so the energy released is \(6 \times 386\) kJ/mol.Total energy released when new bonds form:\[6 \times 386 = 2316 \text{ kJ/mol}\]Now, we apply the relationship for \(\Delta H\):\[\Delta H = \text{energy required to break bonds} - \text{energy released when new bonds form}\]\[\Delta H = 2238 - 2316\]\[\Delta H = -78 \text{ kJ/mol}\]The negative sign indicates that the reaction is exothermic, meaning it releases heat.Therefore, the enthalpy change for the reaction is **-78 kilojoules** when expressed to two significant figures.

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