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Current Question (ID: 8312)

Question:

$\text{The enthalpy of combustion of methane, graphite, and dihydrogen at 298 K are, } -890.3 \text{ kJ mol}^{-1}\text{, } -393.5 \text{ kJ mol}^{-1}\text{, and } -285.8 \text{ kJ mol}^{-1} \text{ respectively. The enthalpy of formation of CH}_4\text{(g) is-}$

Options:

1. $-74.8 \text{ kJ mol}^{-1}$
2. $-52.27 \text{ kJ mol}^{-1}$
3. $+74.8 \text{ kJ mol}^{-1}$
4. $+52.26 \text{ kJ mol}^{-1}$

Solution:

$\text{Hint: Hess's Law of Constant Heat Summation}$ $\text{Write down the desired equation and see from which combination of given reactions we can get the desired reaction.}$ $\text{Explanation:}$ $\text{The following equations are as follows:}$ $(i)\text{ CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(g)}\quad \Delta\text{H} = -890.3 \text{ kJ mol}^{-1}$ $(ii)\text{ C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}\quad \Delta\text{H} = -393.5 \text{ kJ mol}^{-1}$ $(iii)\text{ H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{H}_2\text{O(g)}\quad \Delta\text{H} = -285.8 \text{ kJ mol}^{-1}$ $\text{The desired reaction for the formation of CH}_4\text{(g) is C(s)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}$ $\text{To get this reaction, we need to manipulate the given equations. Let's see how:}$ $\text{We need to get C(s) and H}_2\text{(g) as reactants and CH}_4\text{(g) as product.}$ $\text{From equation (ii), we can get C(s) as a reactant.}$ $\text{From equation (iii), we can get H}_2\text{(g) as a reactant, but we need 2 moles, so we'll multiply equation (iii) by 2.}$ $\text{From equation (i), we can get CH}_4\text{(g) as a reactant, but we need it as a product, so we'll reverse equation (i).}$ $\text{Therefore, the combination is: equation (ii) + 2 × equation (iii) - equation (i)}$ $(ii) + 2(iii) - (i)$ $\text{This gives us:}$ $\text{C(s)} + \text{O}_2\text{(g)} + 2[\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)}] - [\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)}] \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(g)} - [\text{CO}_2\text{(g)} + 2\text{H}_2\text{O(g)}]$ $\text{Simplifying:}$ $\text{C(s)} + \text{O}_2\text{(g)} + 2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} - \text{CH}_4\text{(g)} - 2\text{O}_2\text{(g)} \rightarrow 0$ $\text{Further simplifying:}$ $\text{C(s)} + 2\text{H}_2\text{(g)} - \text{CH}_4\text{(g)} \rightarrow 0$ $\text{Rearranging to get the desired reaction:}$ $\text{C(s)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}$ $\text{Now, let's calculate the enthalpy change:}$ $\Delta_f\text{H}_{\text{CH}_4} = \Delta\text{H}_{(ii)} + 2 \Delta\text{H}_{(iii)} - \Delta\text{H}_{(i)}$ $= [-393.5 + 2(-285.8) - (-890.3)]\text{ kJ mol}^{-1}$ $= [-393.5 - 571.6 + 890.3]\text{ kJ mol}^{-1}$ $= [-965.1 + 890.3]\text{ kJ mol}^{-1}$ $= -74.8\text{ kJ mol}^{-1}$ $\text{Therefore, the enthalpy of formation of CH}_4\text{(g) is } -74.8\text{ kJ mol}^{-1}$

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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
}