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Question:
$\text{The standard enthalpy of the formation of CH}_3\text{OH}_{(l)} \text{ from the following data is:}$ $\text{CH}_3\text{OH}_{(l)} + \frac{3}{2}\text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} + 2\text{H}_2\text{O}_{(l)}; \Delta_r\text{H}^{\circ} = -726 \text{ kJ mol}^{-1}$ $\text{C(s)} + \text{O}_{2(g)} \rightarrow \text{CO}_{2(g)}; \Delta_c\text{H}^{\circ} = -393 \text{ kJ mol}^{-1}$ $\text{H}_{2(g)} + \frac{1}{2}\text{O}_{2(g)} \rightarrow \text{H}_2\text{O}_{(l)}; \Delta_f\text{H}^{\circ} = -286 \text{ kJ mol}^{-1}$
Options:
  • 1. $-239 \text{ kJ mol}^{-1}$
  • 2. $+239 \text{ kJ mol}^{-1}$
  • 3. $-47 \text{ kJ mol}^{-1}$
  • 4. $+47 \text{ kJ mol}^{-1}$
Solution:
$\text{Hint: Write the equation for the formation of CH}_3\text{OH then find standard enthalpy of formation.}$ $\text{Step 1: Note down the given data}$ $(i)\text{ CH}_3 \text{OH}_{(l)} + \frac{3}{2}\text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} + 2\text{H}_2\text{O}_{(l)} ; \Delta_r\text{H}^{\circ} = -726 \text{ kJ mol}^{-1}$ $(ii)\text{ C(s)} + \text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} ; \Delta_c\text{H}^{\circ} = -393 \text{ kJ mol}^{-1}$ $(iii)\text{ H}_{2(g)} + \frac{1}{2}\text{O}_{2(g)} \rightarrow \text{H}_2\text{O}_{(l)} ; \Delta_f\text{H}^{\circ} = -286 \text{ kJ mol}^{-1}$ $\text{Step 2: Write down the equation for the formation of CH}_3\text{OH}_{(l)}$ $\text{The formation of a compound is defined as the reaction in which the compound is formed from its constituent elements in their standard states.}$ $\text{For methanol (CH}_3\text{OH}_{(l)}\text{), the formation reaction is:}$ $(\text{I})\text{ C(s)} + 2\text{H}_{2(g)} + \frac{1}{2}\text{O}_{2(g)} \rightarrow \text{CH}_3 \text{OH}_{(l)}$ $\text{Step 3: Calculate the enthalpy of formation of CH}_3\text{OH}_{(l)}\text{ using the above equations}$ $\text{The reaction (I) can be obtained by following the algebraic calculations as:}$ $\text{Equation (ii)} + 2 \times \text{(Equation (iii))} - \text{equation (i)}$ $\text{Let's verify this combination:}$ $\text{C(s)} + \text{O}_{2(g)} + 2[\text{H}_{2(g)} + \frac{1}{2}\text{O}_{2(g)}] - [\text{CH}_3\text{OH}_{(l)} + \frac{3}{2}\text{O}_{2(g)}] \rightarrow \text{CO}_{2(g)} + 2\text{H}_2\text{O}_{(l)} - [\text{CO}_{2(g)} + 2\text{H}_2\text{O}_{(l)}]$ $\text{Simplifying:}$ $\text{C(s)} + \text{O}_{2(g)} + 2\text{H}_{2(g)} + \text{O}_{2(g)} - \text{CH}_3\text{OH}_{(l)} - \frac{3}{2}\text{O}_{2(g)} \rightarrow 0$ $\text{Further simplifying:}$ $\text{C(s)} + 2\text{H}_{2(g)} + \text{O}_{2(g)} + \text{O}_{2(g)} - \frac{3}{2}\text{O}_{2(g)} - \text{CH}_3\text{OH}_{(l)} \rightarrow 0$ $\text{C(s)} + 2\text{H}_{2(g)} + \frac{1}{2}\text{O}_{2(g)} - \text{CH}_3\text{OH}_{(l)} \rightarrow 0$ $\text{Rearranging to get the desired formation reaction:}$ $\text{C(s)} + 2\text{H}_{2(g)} + \frac{1}{2}\text{O}_{2(g)} \rightarrow \text{CH}_3 \text{OH}_{(l)}$ $\text{Now, let's calculate the enthalpy change:}$ $\text{Thus, } \Delta_f\text{H}^{\circ} [\text{CH}_3 \text{OH (l)}] = \Delta_c\text{H}^{\circ} + 2 \times (\Delta_f\text{H}^{\circ} (\text{H}_2\text{O}_{(l)})) - \Delta_r \text{H}^{\circ}$ $= -393 \text{ kJ mol}^{-1} + 2 \times (-286 \text{ kJ mol}^{-1}) - (-726 \text{ kJ mol}^{-1})$ $= -393 \text{ kJ mol}^{-1} - 572 \text{ kJ mol}^{-1} + 726 \text{ kJ mol}^{-1}$ $= -965 \text{ kJ mol}^{-1} + 726 \text{ kJ mol}^{-1}$ $= -239 \text{ kJ mol}^{-1}$ $\text{Therefore, the standard enthalpy of formation of CH}_3\text{OH}_{(l)} \text{ is } -239 \text{ kJ mol}^{-1}$

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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
}