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If bond energy of H(2),F(2) and HF are i...

If bond energy of `H_(2),F_(2)` and `HF` are in the ratio `2:1:3` and `DeltaH_(a)(H_(2))=400kJ//mol`. Then `DeltaH_(f)(HF)` is `:`

A

`0 kJ //mol`

B

`-600kJ //mol`

C

`-1200kJ//mol`

D

None of these

Text Solution

AI Generated Solution

The correct Answer is:
To find the heat of formation of HF (ΔH_f(HF)), we will follow these steps: ### Step 1: Define the Bond Energies Given the bond energies of H₂, F₂, and HF are in the ratio 2:1:3, we can express them in terms of a variable \( X \): - Bond energy of H₂ = \( 2X \) - Bond energy of F₂ = \( X \) - Bond energy of HF = \( 3X \) ### Step 2: Use the Given Information We know that the heat of dissociation of H₂ (ΔH_a(H₂)) is 400 kJ/mol. Since the bond energy of H₂ is \( 2X \), we can set up the equation: \[ 2X = 400 \text{ kJ/mol} \] From this, we can solve for \( X \): \[ X = \frac{400}{2} = 200 \text{ kJ/mol} \] ### Step 3: Calculate the Bond Energies Now we can calculate the bond energies for each molecule: - Bond energy of H₂ = \( 2X = 400 \text{ kJ/mol} \) - Bond energy of F₂ = \( X = 200 \text{ kJ/mol} \) - Bond energy of HF = \( 3X = 3 \times 200 = 600 \text{ kJ/mol} \) ### Step 4: Write the Formation Reaction The formation of HF from its elements can be represented as: \[ \frac{1}{2} \text{H}_2 + \frac{1}{2} \text{F}_2 \rightarrow \text{HF} \] ### Step 5: Apply the Heat of Formation Formula The heat of formation (ΔH_f) can be calculated using the formula: \[ \Delta H_f(HF) = \text{(Bond energy of reactants)} - \text{(Bond energy of products)} \] Substituting the bond energies into the equation: \[ \Delta H_f(HF) = \left( \frac{1}{2} \times 400 + \frac{1}{2} \times 200 \right) - 600 \] ### Step 6: Calculate the Values Calculating the values: 1. For H₂: \( \frac{1}{2} \times 400 = 200 \) 2. For F₂: \( \frac{1}{2} \times 200 = 100 \) Now substituting these into the equation: \[ \Delta H_f(HF) = (200 + 100) - 600 \] \[ \Delta H_f(HF) = 300 - 600 \] \[ \Delta H_f(HF) = -300 \text{ kJ/mol} \] ### Final Answer Thus, the heat of formation of HF is: \[ \Delta H_f(HF) = -300 \text{ kJ/mol} \] ---

To find the heat of formation of HF (ΔH_f(HF)), we will follow these steps: ### Step 1: Define the Bond Energies Given the bond energies of H₂, F₂, and HF are in the ratio 2:1:3, we can express them in terms of a variable \( X \): - Bond energy of H₂ = \( 2X \) - Bond energy of F₂ = \( X \) - Bond energy of HF = \( 3X \) ...
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