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Question:
$\text{The enthalpy of formation of CO}_{(g)}\text{, CO}_{2(g)}\text{, N}_2\text{O}_{(g)}\text{, and N}_2\text{O}_{4(g)}\text{ are } -110\text{ kJ mol}^{-1}\text{, } -393\text{ kJ mol}^{-1}\text{, } 81\text{ kJ mol}^{-1}\text{, and } 9.7\text{ kJ mol}^{-1}\text{ respectively.}$ $\text{The value of } (\Delta)_r H \text{ for the reaction would be:}$ $\text{N}_2\text{O}_{4(g)} + 3\text{(CO)}_{(g)} \rightarrow \text{N}_2\text{O}_{(g)} + 3\text{(CO)}_{2(g)}$
Options:
  • 1. $-777.7\text{ kJ mol}^{-1}$
  • 2. $+777.7\text{ kJ mol}^{-1}$
  • 3. $+824.9\text{ kJ mol}^{-1}$
  • 4. $-345.4\text{ kJ mol}^{-1}$
Solution:
$\text{Hint: } \Delta \text{ H}_r = \sum \Delta\text{H}_f\text{(Products)} - \sum \Delta\text{H}_f\text{(Reactants)}$ $\text{Step 1: Write down the given data}$ $\Delta_f \text{H(CO}_{(g)}\text{)} = -110\text{ kJ mol}^{-1}$ $\Delta_f \text{H(CO}_{2(g)}\text{)} = -393\text{ kJ mol}^{-1}$ $\Delta_f \text{H(N}_2\text{O}_{(g)}\text{)} = 81\text{ kJ mol}^{-1}$ $\Delta_f \text{H(N}_2\text{O}_{4(g)}\text{)} = 9.7\text{ kJ mol}^{-1}$ $\text{Step 2: Calculate } \Delta_r\text{H for the given reaction}$ $\text{Reaction: N}_2\text{O}_{4(g)}\text{ + 3 CO}_{(g)}\text{ } \rightarrow \text{ N}_2\text{O}_{(g)}\text{ + 3 CO}_{2(g)}$ $\Delta_r\text{H} = [\Delta_f\text{H (N}_2\text{O}_{(g)}\text{)} + 3 \Delta_f\text{H (CO}_{2(g)}\text{)}] - [\Delta_f\text{H (N}_2\text{O}_{4(g)}\text{)} + 3 \Delta_f\text{H (CO}_{(g)}\text{)}]$ $= \{81 + 3 (-393) - [9.7 + 3 (-110)]\} = \{81 - 1179 - [9.7 - 330]\} = \{-1098 - [-320.3]\} = -777.7\text{ kJ mol}^{-1}$ $\text{Hence, the value of } \Delta_r\text{H for the reaction is } -777.7\text{ kJ mol}^{-1}$ $\text{The negative value of enthalpy change indicates that this is an exothermic reaction, releasing energy in the form of heat.}$

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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
}