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1 Le Chatelier's principle states that if a change in conditions (such as temperature, pressure, or concentration of reactants) is applied to a system at equilibrium, the system will shift in a

Physics

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8.5

300 K\text{300 K}

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8.1 Le Chatelier's principle states that if a change in conditions (such as temperature, pressure, or concentration of reactants) is applied to a system at equilibrium, the system will shift in a direction that counteracts the change to re-establish equilibrium.

8.2 At 700 K, the equilibrium constant (KcK_c) is 1.20×1041.20 \times 10^4. This is a very large value, indicating that the equilibrium lies far to the right, favoring the formation of products. The product, NO2 (g)\text{NO}_2\text{ (g)}, is a brown gas. Therefore, the appearance of the gas jar at 700 K will be brown.

Part 8.3

To determine if the reaction is exothermic or endothermic, we observe the trend of KcK_c with temperature from the table

  • As the temperature increases from 300 K to 700 K, the value of KcK_c significantly increases (from 1.00×10−11.00 \times 10^{-1} to 1.20×1041.20 \times 10^4).
  • An increase in KcK_c with increasing temperature means that the equilibrium shifts to favor the products (right side of the reaction).
  • According to Le Chatelier's principle, if increasing the temperature shifts the equilibrium to the product side, the forward reaction must be endothermic (absorbs heat). This is because the system tries to absorb the added heat by favoring the endothermic direction.

Part 8.4

Two ways, other than temperature change, that can increase the rate of the forward reaction at 500 K are

  • Increasing the concentration of the reactant, N2O4 (g)\text{N}_2\text{O}_4\text{ (g)}.
  • Adding a suitable catalyst.

Part 8.5

300 K\text{300 K}

The reaction is: N2O4 (g)⇌2NO2 (g)\text{N}_2\text{O}_4\text{ (g)} \rightleftharpoons 2\text{NO}_2\text{ (g)}

  1. Calculate the initial moles of N2O4\text{N}_2\text{O}_4.

    The molar mass of N2O4\text{N}_2\text{O}_4 is 2(14.01)+4(16.00)=28.02+64.00=92.02 g/mol2(14.01) + 4(16.00) = 28.02 + 64.00 = 92.02 \text{ g/mol}.
    Initial moles of N2O4=massmolar mass=84.64 g92.02 g/mol≈0.920 mol\text{N}_2\text{O}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{84.64 \text{ g}}{92.02 \text{ g/mol}} \approx 0.920 \text{ mol}.

  2. Calculate the moles of N2O4\text{N}_2\text{O}_4 that decomposed.

    20.7%20.7\% of N2O4\text{N}_2\text{O}_4 decomposed.
    Moles decomposed =0.207×0.920 mol=0.19044 mol= 0.207 \times 0.920 \text{ mol} = 0.19044 \text{ mol}.

  3. Set up an ICE (Initial, Change, Equilibrium) table for moles.

    SpeciesN2O4\text{N}_2\text{O}_4 (mol)2NO22\text{NO}_2 (mol)
    Initial0.9200
    Change−0.19044-0.19044+2×0.19044=+0.38088+2 \times 0.19044 = +0.38088
    Equil.0.920−0.19044=0.729560.920 - 0.19044 = 0.729560.38088
  4. Calculate the equilibrium concentrations.

    The volume of the container is 2 dm32 \text{ dm}^3.
    [N2O4]eq=0.72956 mol2 dm3=0.36478 mol/dm3[\text{N}_2\text{O}_4]_{\text{eq}} = \frac{0.72956 \text{ mol}}{2 \text{ dm}^3} = 0.36478 \text{ mol/dm}^3.
    [NO2]eq=0.38088 mol2 dm3=0.19044 mol/dm3[\text{NO}_2]_{\text{eq}} = \frac{0.38088 \text{ mol}}{2 \text{ dm}^3} = 0.19044 \text{ mol/dm}^3.

  5. Calculate the equilibrium constant, KcK_c.

    Kc=[NO2]2[N2O4]K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} Kc=(0.19044)20.36478K_c = \frac{(0.19044)^2}{0.36478} Kc=0.03626743360.36478K_c = \frac{0.0362674336}{0.36478} Kc≈0.09942K_c \approx 0.09942
  6. Determine the temperature T from the table.

    The calculated Kc≈0.09942K_c \approx 0.09942 is approximately 1.00×10−11.00 \times 10^{-1}.
    From the table, a KcK_c value of 1.00×10−11.00 \times 10^{-1} corresponds to a temperature of 300 K300 \text{ K}.
    Therefore, the temperature T at which the reaction took place is 300 K\boxed{\text{300 K}}.

Part 8.6

The effect of adding more N2O4 (g)\text{N}_2\text{O}_4\text{ (g)} on the value of the equilibrium constant, KcK_c, is that it *REMAINS THE SAME

Reason: The equilibrium constant (KcK_c) is a constant value for a specific reaction at a given temperature. It does not change with changes in the concentration of reactants or products. Adding more N2O4\text{N}_2\text{O}_4 will cause the equilibrium to shift to the right to consume the added reactant, but the ratio of products to reactants at equilibrium (i.e., KcK_c) will remain unchanged as long as the temperature is constant.

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