For N2O5( in CCl4)----2NO2+1/2O2,K=6*10-4at 350kand K=1.2*10-3AT 360K,THEN WHEN TEAMPERATURE IS CHANGED TO 380kVALUE OF k in sec-1

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To determine the equilibrium constant \( K \) for the reaction of dinitrogen pentoxide (\( N_2O_5 \)) decomposing into nitrogen dioxide (\( NO_2 \)) and oxygen (\( O_2 \)) at a temperature of 380 K, we can utilize the Van 't Hoff equation. This equation relates the change in the equilibrium constant with temperature to the enthalpy change of the reaction. Let's break this down step by step.

Understanding the Reaction and Constants

The reaction we are considering is:

• Reaction: \( N_2O_5 \) (in \( CCl_4 \)) ⇌ \( 2NO_2 + \frac{1}{2}O_2 \)

We have the equilibrium constants at two different temperatures:

• K at 350 K: \( 6 \times 10^{-4} \)
• K at 360 K: \( 1.2 \times 10^{-3} \)

Applying the Van 't Hoff Equation

The Van 't Hoff equation is given by:

\( \ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H^\circ}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \)

Where:

• \( K_1 \) and \( K_2 \) are the equilibrium constants at temperatures \( T_1 \) and \( T_2 \), respectively.
• \( \Delta H^\circ \) is the standard enthalpy change of the reaction.
• \( R \) is the universal gas constant, approximately \( 8.314 \, J/(mol \cdot K) \).

Calculating the Enthalpy Change

First, we can use the values at 350 K and 360 K to find \( \Delta H^\circ \). Let's denote:

• \( K_1 = 6 \times 10^{-4} \) at \( T_1 = 350 \, K \)
• \( K_2 = 1.2 \times 10^{-3} \) at \( T_2 = 360 \, K \)

Plugging these values into the Van 't Hoff equation:

\( \ln\left(\frac{1.2 \times 10^{-3}}{6 \times 10^{-4}}\right) = -\frac{\Delta H^\circ}{8.314} \left(\frac{1}{360} - \frac{1}{350}\right) \)

Calculating the Left Side

Calculating the natural logarithm:

\( \ln\left(\frac{1.2 \times 10^{-3}}{6 \times 10^{-4}}\right) = \ln(2) \approx 0.693 \)

Calculating the Right Side

Now, calculate the temperature difference:

\( \frac{1}{360} - \frac{1}{350} = \frac{350 - 360}{360 \times 350} = -\frac{10}{126000} \approx -7.9365 \times 10^{-5} \)

Now substituting these values into the equation:

\( 0.693 = -\frac{\Delta H^\circ}{8.314} \times (-7.9365 \times 10^{-5}) \)

Solving for \( \Delta H^\circ \):

\( \Delta H^\circ = \frac{0.693 \times 8.314}{7.9365 \times 10^{-5}} \approx 7.1 \times 10^{4} \, J/mol \)

Finding \( K \) at 380 K

Now that we have \( \Delta H^\circ \), we can find \( K \) at 380 K using the Van 't Hoff equation again:

\( \ln\left(\frac{K_{380}}{1.2 \times 10^{-3}}\right) = -\frac{7.1 \times 10^{4}}{8.314} \left(\frac{1}{380} - \frac{1}{360}\right) \)

Calculating the temperature difference:

\( \frac{1}{380} - \frac{1}{360} = \frac{360 - 380}{380 \times 360} = -\frac{20}{136800} \approx -1.459 \times 10^{-4} \)

Substituting back into the equation:

\( \ln\left(\frac{K_{380}}{1.2 \times 10^{-3}}\right) = -\frac{7.1 \times 10^{4}}{8.314} \times (-1.459 \times 10^{-4}) \)

Calculating the right side gives us:

\( \ln\left(\frac{K_{380}}{1.2 \times 10^{-3}}\right) \approx 1.24 \)

Exponentiating both sides:

\( \frac{K_{380}}{1.2 \times 10^{-3}} \approx e^{1.24} \approx 3.46 \)

Finally, solving for \( K_{380} \):

\( K_{380} \approx 3.46 \times 1.2 \times 10^{-3} \approx 4.15 \times 10^{-3} \)

Final Result

Thus, the equilibrium constant \( K \) at 380 K is approximately \( 4.15 \times 10^{-3} \). This value indicates how the equilibrium position shifts with temperature, reflecting the endothermic nature of the decomposition reaction.