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Current Question (ID: 8428)

Question:

$\text{For the reaction: FeO}_{(s)} + \text{CO}_{(g)} \rightleftharpoons \text{Fe}_{(s)} + \text{CO}_2\text{(g)}, K_p = 0.265 \text{ at } 1050 \text{ K. If the initial partial pressures are } p_{\text{CO}} = 1.4 \text{ atm and } p_{\text{CO}_2} = 0.80 \text{ atm, the partial pressure of CO}_2 \text{ at equilibrium at } 1050 \text{ K would be:}$

Options:

1. $4.61 \text{ atm}$
2. $1.74 \text{ atm}$
3. $0.46 \text{ atm}$
4. $0.17 \text{ atm}$

Solution:

$\text{Hint: If } Q_p > K_p\text{, the reaction will proceed in the backward direction.}$ $\text{Explanation:}$ $\text{Step 1: Calculate } Q_p$ $\text{For the reaction:}$ $\text{FeO}_{(s)} + \text{CO}_{(g)} \rightleftharpoons \text{Fe}_{(s)} + \text{CO}_2\text{(g)}$ $\text{Initial } p \quad \quad \quad 1.4\text{atm} \quad \quad \quad 0.80\text{atm}$ $Q_p = \frac{p_{\text{CO}_2}}{p_{\text{CO}}} = \frac{0.80}{1.4}$ $Q_p = 0.571 \text{ and } K_p = 0.265 \text{ (given)}$ $\text{If } Q_p > K_p\text{, the reaction will proceed in the backward direction.}$ $\text{Step 2: Find partial pressure of CO}_2$ $\text{Therefore, we can say that the pressure of CO will increase while the pressure of CO}_2 \text{ will decrease.}$ $\text{Now, let the increase in pressure of CO = decrease in pressure of CO}_2 \text{ be } p.$ $\text{Then, we can write,}$ $K_p = \frac{p_{\text{CO}_2}}{p_{\text{CO}}}$ $\Rightarrow 0.265 = \frac{0.80-p}{1.4+p}$ $\Rightarrow p = 0.339\text{atm}$ $\text{Therefore, equilibrium partial pressure of CO}_2, p_{\text{CO}_2} = 0.80 - 0.339 = 0.46 \text{ atm}$

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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
}