n₂(g) + 3f₂(g) → 2nf₃(g) δh°ᵣₓₙ = -264 kj/molᵣₓₙ\nthe bond enthalpies of the n≡n and f - f bonds are 945…

n₂(g) + 3f₂(g) → 2nf₃(g) δh°ᵣₓₙ = -264 kj/molᵣₓₙ\nthe bond enthalpies of the n≡n and f - f bonds are 945 kj/mol and 159 kj/mol, respectively. based on the value of δh°ᵣₓₙ for the reaction represented above, what is the average bond enthalpy (in kj/mol) of a n - f bond in nf₃?\n\n kj/mol

n₂(g) + 3f₂(g) → 2nf₃(g) δh°ᵣₓₙ = -264 kj/molᵣₓₙ\nthe bond enthalpies of the n≡n and f - f bonds are 945 kj/mol and 159 kj/mol, respectively. based on the value of δh°ᵣₓₙ for the reaction represented above, what is the average bond enthalpy (in kj/mol) of a n - f bond in nf₃?\n\n kj/mol

Answer

Explanation:

Step1: Write the formula for $\Delta H^{\circ}_{rxn}$

$\Delta H^{\circ}_{rxn}=\sum$ (Bond enthalpies of reactants) - $\sum$ (Bond enthalpies of products)

Step2: Calculate bond - enthalpies of reactants

For the reactants, there is 1 $N\equiv N$ bond and 3 $F - F$ bonds. The bond - enthalpy of reactants is $1\times945\ kJ/mol+3\times159\ kJ/mol=945 + 477=1422\ kJ/mol$

Step3: Calculate bond - enthalpies of products

In $2NF_3$, there are $2\times3 = 6$ $N - F$ bonds. Let the bond enthalpy of $N - F$ be $x\ kJ/mol$. So the bond - enthalpy of products is $6x\ kJ/mol$

Step4: Substitute values into $\Delta H^{\circ}_{rxn}$ formula

We know that $\Delta H^{\circ}_{rxn}=- 264\ kJ/mol$. $-264\ kJ/mol=1422\ kJ/mol-6x\ kJ/mol$

Step5: Solve for $x$

First, rearrange the equation: $6x=1422 + 264$ $6x=1686$ $x=\frac{1686}{6}=281\ kJ/mol$

Answer:

$281$