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Current Question (ID: 8317)

Question:

$\text{At standard conditions, if the change in the enthalpy for the following reaction is } -109 \text{ kJ mol}^{-1}$ $\text{H}_{2(g)}+\text{Br}_{2(g)}\rightarrow 2\text{HBr}_{(g)} \text{ and the bond energy of H}_2 \text{ and Br}_2 \text{ is } 435 \text{ kJ mol}^{-1} \text{ and } 192 \text{ kJ mol}^{-1} \text{ respectively, what is the bond energy (in kJ mol}^{-1}\text{) of HBr?}$

Options:

1. $368$
2. $736$
3. $518$
4. $259$

Solution:

$\text{Hint} = \Delta\text{H} = \sum(\text{B.E})_{\text{Reactants}} - \sum(\text{B.E})_{\text{Products}}$ $\text{Step 1: Identify the reaction and the given data}$ $\text{The reaction is: H}_{2(g)} + \text{Br}_{2(g)} \rightarrow 2\text{HBr}_{(g)}$ $\text{Given:}$ $\Delta\text{H} = -109 \text{ kJ mol}^{-1}$ $\text{Bond energy of H}_2 = 435 \text{ kJ mol}^{-1}$ $\text{Bond energy of Br}_2 = 192 \text{ kJ mol}^{-1}$ $\text{Step 2: Apply the relationship between enthalpy change and bond energies}$ $\text{The enthalpy change for a reaction can be calculated as:}$ $\Delta\text{H} = \sum(\text{Bond energies broken}) - \sum(\text{Bond energies formed})$ $\text{Or alternatively:}$ $\Delta\text{H} = \sum(\text{B.E})_{\text{Reactants}} - \sum(\text{B.E})_{\text{Products}}$ $\text{In this reaction:}$ $\text{Bonds broken: one H-H bond and one Br-Br bond}$ $\text{Bonds formed: two H-Br bonds}$ $\text{Step 3: Set up the equation and solve for the bond energy of HBr}$ $-109 = [\text{B.E}_{(\text{H-H})} + \text{B.E}_{(\text{Br-Br})}] - [2 \times \text{B.E}_{(\text{H-Br})}]$ $-109 = 435 + 192 - 2 \times \text{B.E}_{(\text{H-Br})}$ $-109 = 627 - 2 \times \text{B.E}_{(\text{H-Br})}$ $2 \times \text{B.E}_{(\text{H-Br})} = 627 + 109$ $2 \times \text{B.E}_{(\text{H-Br})} = 736$ $\text{B.E}_{(\text{H-Br})} = \frac{736}{2} = 368 \text{ kJ mol}^{-1}$ $\text{Therefore, the bond energy of HBr is } 368 \text{ kJ mol}^{-1}$

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{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}