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question 17 (1 point) saved use the following information to answer the…

Question

question 17 (1 point) saved
use the following information to answer the next question.
nitric oxide (no(g)) is a colourless gas. the main
sources of nitric oxide are power plants and automobile
engines. the gas is produced by many methods. one of
these methods is given below:
so₂(g) + no₂(g) ↔ no(g) + so₃(g)
suppose that at 700 k, the initial concentration of so₂(g) is 0.25 mol/l and that of
no₂(g) is also 0.25 mol/l. if the equilibrium concentration of so₂(g) is 0.080 mol/l,
the value of equilibrium constant at that temperature will be
0.037
0.22
2.7
4.0
4.5

Explanation:

Step1: Determine the change in concentration of \( \text{SO}_2 \)

The initial concentration of \( \text{SO}_2 \) is \( 0.25 \, \text{mol/L} \), and the equilibrium concentration is \( 0.080 \, \text{mol/L} \). The change in concentration (\( \Delta[\text{SO}_2] \)) is \( 0.25 - 0.080 = 0.17 \, \text{mol/L} \).

Step2: Determine equilibrium concentrations of all species

From the reaction \( \text{SO}_2(\text{g}) + \text{NO}_2(\text{g})
ightleftharpoons \text{NO}(\text{g}) + \text{SO}_3(\text{g}) \), the stoichiometry is 1:1:1:1. So:

  • \( [\text{SO}_2]_{\text{eq}} = 0.080 \, \text{mol/L} \)
  • \( [\text{NO}_2]_{\text{eq}} = 0.25 - 0.17 = 0.080 \, \text{mol/L} \) (since change in \( \text{NO}_2 \) is same as \( \text{SO}_2 \))
  • \( [\text{NO}]_{\text{eq}} = 0 + 0.17 = 0.17 \, \text{mol/L} \) (initial \( \text{NO} \) is 0)
  • \( [\text{SO}_3]_{\text{eq}} = 0 + 0.17 = 0.17 \, \text{mol/L} \) (initial \( \text{SO}_3 \) is 0)

Step3: Calculate the equilibrium constant (\( K_c \))

The formula for \( K_c \) is \( K_c = \frac{[\text{NO}][\text{SO}_3]}{[\text{SO}_2][\text{NO}_2]} \).
Substitute the equilibrium concentrations:
\( K_c = \frac{(0.17)(0.17)}{(0.080)(0.080)} = \frac{0.0289}{0.0064} \approx 4.5156 \)? Wait, no, wait, maybe I made a mistake. Wait, wait, initial concentrations: \( \text{SO}_2 \) and \( \text{NO}_2 \) are both 0.25. Change is \( x = 0.25 - 0.08 = 0.17 \). So \( [\text{NO}] = x = 0.17 \), \( [\text{SO}_3] = x = 0.17 \), \( [\text{SO}_2] = 0.25 - x = 0.08 \), \( [\text{NO}_2] = 0.25 - x = 0.08 \). Then \( K_c = \frac{(0.17)(0.17)}{(0.08)(0.08)} = \frac{0.0289}{0.0064} \approx 4.5156 \)? But the option with 4.5 is there. Wait, maybe my calculation of change is wrong? Wait, wait, the equilibrium concentration of \( \text{SO}_2 \) is 0.080, initial is 0.25, so change is 0.25 - 0.08 = 0.17. So \( \text{NO}_2 \) also changes by 0.17, so equilibrium \( \text{NO}_2 \) is 0.25 - 0.17 = 0.08. Then \( \text{NO} \) and \( \text{SO}_3 \) are 0.17 each. Then \( K_c = (0.17 0.17)/(0.08 0.08) = (0.0289)/(0.0064) ≈ 4.5156 \), which is approximately 4.5. But the selected option is 2.7. Wait, maybe I messed up the initial assumption? Wait, maybe the initial concentration of \( \text{NO} \) or \( \text{SO}_3 \) is not zero? Wait, the problem says "the gas is produced by many methods" but the reaction given is the method, so initial concentrations of \( \text{NO} \) and \( \text{SO}_3 \) are zero. Wait, maybe the question has a typo, or my calculation is wrong. Wait, let's recalculate: \( 0.17 * 0.17 = 0.0289 \), \( 0.08 * 0.08 = 0.0064 \), \( 0.0289 / 0.0064 ≈ 4.5156 \), which is ~4.5. So the correct answer should be 4.5? But the selected option is 2.7. Wait, maybe I made a mistake in the change. Wait, initial \( \text{SO}_2 \) is 0.25, equilibrium is 0.08, so change is 0.17. Then \( \text{NO}_2 \) also changes by 0.17, so equilibrium \( \text{NO}_2 \) is 0.25 - 0.17 = 0.08. Then \( \text{NO} \) and \( \text{SO}_3 \) are 0.17. Then \( K_c = (0.17)(0.17)/(0.08)(0.08) = 0.0289 / 0.0064 ≈ 4.5 \). So the answer should be 4.5, which is option E (if options are 0.037, 0.22, 2.7, 4.0, 4.5). Wait, the options are 0.037, 0.22, 2.7, 4.0, 4.5. So 4.5 is the correct one. But the selected option in the image is 2.7, maybe that's a mistake. But according to calculation, it's 4.5. Wait, maybe I messed up the reaction? Wait, the reaction is \( \text{SO}_2 + \text{NO}_2
ightleftharpoons \text{NO} + \text{SO}_3 \). So \( K_c = [\text{NO}][\text{SO}_3]/([\text{SO}_2][\text{NO}_2]) \). Yes. So with the numbers, it's (0.170.17)/(0.080.08) ≈ 4.5. So t…

Answer:

4.5 (the option with 4.5, e.g., if options are labeled as E. 4.5, then E. 4.5)