Certainly! Let's work through the problem step-by-step to determine the mean enthalpy of formation (∆Hv) for one mole of ammonia (NH3) from its elements within the specified temperature range.
### Given Data:
We have the following given values:
- Temperature 1 (T1): 673 K
- Temperature 2 (T2): 773 K
- Equilibrium constant at 673 K (Kp1): 1.64 x 10
- Equilibrium constant at 773 K (Kp2): 11.44 x 10^5
- Universal gas constant (R): 8.314 J/(mol·K)
### Step-by-Step Solution:
1. Write the Van't Hoff equation:
The Van't Hoff equation relates the change in the equilibrium constant with temperature and the change in enthalpy:
[tex]\[
\ln\left(\frac{K_{p2}}{K_{p1}}\right) = -\frac{\Delta H_v}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)
\][/tex]
2. Calculate the ratio of equilibrium constants:
[tex]\[
\ln\left(\frac{K_{p2}}{K_{p1}}\right)
\][/tex]
Given:
[tex]\[
K_{p1} = 1.64 \times 10
\][/tex]
[tex]\[
K_{p2} = 11.44 \times 10^5
\][/tex]
Therefore:
[tex]\[
\ln\left(\frac{K_{p2}}{K_{p1}}\right) \approx 11.152760116091727
\][/tex]
3. Calculate the inverse of the temperatures:
[tex]\[
\frac{1}{T_1} = \frac{1}{673} \approx 0.0014858841010401188 \text{ K}^{-1}
\][/tex]
[tex]\[
\frac{1}{T_2} = \frac{1}{773} \approx 0.00129366106080207 \text{ K}^{-1}
\][/tex]
4. Substitute all known quantities into the Van't Hoff equation and solve for ∆Hv:
[tex]\[
11.152760116091727 = -\frac{\Delta H_v}{8.314} \left(0.00129366106080207 - 0.0014858841010401188 \right)
\][/tex]
Calculate the difference in inverse temperatures:
[tex]\[
0.00129366106080207 - 0.0014858841010401188 \approx -0.0001922230402380488 \text{ K}^{-1}
\][/tex]
5. Rearrange to solve for ∆Hv:
[tex]\[
\Delta H_v = -11.152760116091727 \times 8.314 \times \left(-0.0001922230402380488\right)
\][/tex]
[tex]\[
\Delta H_v \approx 482377.3856159868 \text{ J/mol}
\][/tex]
### Conclusion:
The mean enthalpy of formation (∆Hv) for one mole of ammonia from its elements in the temperature range of 673 K to 773 K is approximately 482,377.39 J/mol or 482.38 kJ/mol.
