Calculating Standard Heat Of Reaction: N2 + 3H2 -> 2NH3

Hey guys! Let's dive into calculating the standard heat of reaction for the synthesis of ammonia ($N_2(g) + 3H_2(g) 2NH_3(g)$) at two different temperatures: 25°C and 300°C. This is a classic problem in chemical thermodynamics, and understanding how to tackle it is super important. We'll break it down step by step, making sure it's easy to follow. So, grab your calculators, and let's get started!

Understanding Standard Heat of Reaction

Before we jump into the calculations, let's quickly recap what the standard heat of reaction ($\Delta H^{\circ}$) actually means. Simply put, it's the change in enthalpy when a reaction occurs under standard conditions (usually 298 K or 25°C and 1 atm pressure). It tells us whether a reaction releases heat (exothermic, $\Delta H^{\circ} < 0$) or absorbs heat (endothermic, $\Delta H^{\circ} > 0$).

In our case, we're interested in the reaction:

$N_2(g) + 3H_2(g) 2NH_3(g)$

We're given the standard enthalpy of formation ($\Delta H_{298}^{\circ}$) for each species, as well as coefficients A, B, C, and D that will help us calculate the heat capacities at different temperatures. Remember, the standard enthalpy of formation is the change in enthalpy when one mole of a compound is formed from its elements in their standard states.

Step 1: Calculating $\Delta H^{\circ}_{298}$ (at 25°C)

First, we need to calculate the standard heat of reaction at 25°C (298 K). We can use the following formula:

$\Delta H^{\circ}_{298} = \sum [n \cdot \Delta H^{\circ}_{f,products}] - \sum [n \cdot \Delta H^{\circ}_{f,reactants}]$

where:

• $n$ is the stoichiometric coefficient of each species in the balanced chemical equation.
• $\Delta H^{\circ}_{f}$ is the standard enthalpy of formation of each species.

From the table provided (which you'll need to fill in with the actual values), we have:

• $\Delta H^{\circ}_{f,NH_3}$ (Ammonia)
• $\Delta H^{\circ}_{f,N_2}$ (Nitrogen) = 0 kJ/mol (since it's an element in its standard state)
• $\Delta H^{\circ}_{f,H_2}$ (Hydrogen) = 0 kJ/mol (since it's an element in its standard state)

Let's plug in the values:

$\Delta H^{\circ}_{298} = [2 \cdot \Delta H^{\circ}_{f,NH_3}] - [1 \cdot \Delta H^{\circ}_{f,N_2} + 3 \cdot \Delta H^{\circ}_{f,H_2}]$

$\Delta H^{\circ}_{298} = [2 \cdot \Delta H^{\circ}_{f,NH_3}] - [1 \cdot 0 + 3 \cdot 0]$

$\Delta H^{\circ}_{298} = 2 \cdot \Delta H^{\circ}_{f,NH_3}$

So, the standard heat of reaction at 25°C is simply twice the standard enthalpy of formation of ammonia. Make sure you use the correct value for $\Delta H^{\circ}_{f,NH_3}$ from your table. The more accurate the values, the better the final answer. We need to make sure to get this right!

Step 2: Calculating $\Delta H^{\circ}_{300}$ (at 300°C)

Now comes the slightly trickier part: calculating the standard heat of reaction at 300°C (573 K). Since the temperature has changed, we need to account for the change in heat capacities of the reactants and products. We'll use Kirchhoff's Law, which states:

$\Delta H^{\circ}_{T_2} = \Delta H^{\circ}_{T_1} + \int_{T_1}^{T_2} \Delta C_p dT$

where:

• $\Delta H^{\circ}_{T_2}$ is the standard heat of reaction at temperature $T_2$ (300°C or 573 K).
• $\Delta H^{\circ}_{T_1}$ is the standard heat of reaction at temperature $T_1$ (25°C or 298 K), which we calculated in the previous step.
• $\Delta C_p$ is the change in heat capacity at constant pressure.
• $dT$ is the differential change in temperature.

Calculating $\Delta C_p$

$\Delta C_p$ is calculated similarly to $\Delta H^{\circ}$:

$\Delta C_p = \sum [n \cdot C_{p,products}] - \sum [n \cdot C_{p,reactants}]$

And the heat capacity for each species can be expressed as a function of temperature using the given coefficients A, B, C, and D:

$C_p = A + BT + CT^2 + DT^{-2}$

So, for our reaction:

$\Delta C_p = [2 \cdot C_{p,NH_3}] - [1 \cdot C_{p,N_2} + 3 \cdot C_{p,H_2}]$

Now, we need to substitute the expression for $C_p$ for each species:

$\Delta C_p = 2[A_{NH_3} + B_{NH_3}T + C_{NH_3}T^2 + D_{NH_3}T^{-2}] - [A_{N_2} + B_{N_2}T + C_{N_2}T^2 + D_{N_2}T^{-2}] - 3[A_{H_2} + B_{H_2}T + C_{H_2}T^2 + D_{H_2}T^{-2}]$

Integrating $\Delta C_p$ with Respect to Temperature

Now we have a rather long expression for $\Delta C_p$ in terms of temperature. We need to integrate this expression from $T_1$ (298 K) to $T_2$ (573 K):

$\int_{T_1}^{T_2} \Delta C_p dT = \int_{298}^{573} \{2[A_{NH_3} + B_{NH_3}T + C_{NH_3}T^2 + D_{NH_3}T^{-2}] - [A_{N_2} + B_{N_2}T + C_{N_2}T^2 + D_{N_2}T^{-2}] - 3[A_{H_2} + B_{H_2}T + C_{H_2}T^2 + D_{H_2}T^{-2}]\} dT$

This looks daunting, but it's just a polynomial integration. We can integrate term by term. The key here is to be meticulous with your algebra and keep track of all the terms.

The integral of $A$ with respect to $T$ is $AT$. The integral of $BT$ with respect to $T$ is $\frac{1}{2}BT^2$. The integral of $CT^2$ with respect to $T$ is $\frac{1}{3}CT^3$. The integral of $DT^{-2}$ with respect to $T$ is $-DT^{-1}$.

So, the result of the integration will look something like this (I'm omitting the full expression for brevity, but you'll need to plug in all the A, B, C, and D values):

$\int_{298}^{573} \Delta C_p dT = [2A_{NH_3}T + B_{NH_3}T^2 + (2/3)C_{NH_3}T^3 - 2D_{NH_3}T^{-1}] - [A_{N_2}T + (1/2)B_{N_2}T^2 + (1/3)C_{N_2}T^3 - D_{N_2}T^{-1}] - 3[A_{H_2}T + (3/2)B_{H_2}T^2 + C_{H_2}T^3 - 3D_{H_2}T^{-1}] evaluated from 298 to 573

After evaluating the integral at the upper and lower limits (573 K and 298 K) and subtracting, you'll get a numerical value for $\int_{298}^{573} \Delta C_p dT$.

Final Calculation of $\Delta H^{\circ}_{300}$

Finally, we can plug everything back into Kirchhoff's Law:

$\Delta H^{\circ}_{573} = \Delta H^{\circ}_{298} + \int_{298}^{573} \Delta C_p dT$

We already calculated $\Delta H^{\circ}_{298}$ in Step 1, and we just calculated the integral in the previous step. Add them together, and you'll get the standard heat of reaction at 300°C (573 K).

Summary and Important Considerations

Okay, let's recap what we've done:

1. Calculated $\Delta H^{\circ}_{298}$: Used the standard enthalpies of formation to find the standard heat of reaction at 25°C.
2. Calculated $\Delta C_p$: Found the change in heat capacity as a function of temperature using the given coefficients.
3. Integrated $\Delta C_p$: Integrated the $\Delta C_p$ expression with respect to temperature from 298 K to 573 K.
4. Applied Kirchhoff's Law: Used Kirchhoff's Law to find the standard heat of reaction at 300°C.

Important Considerations:

• Units: Make sure all your units are consistent (kJ/mol, K, etc.). A small error in units can throw off your entire calculation.
• Sign Conventions: Pay close attention to the signs of $\Delta H^{\circ}_{f}$ and the coefficients A, B, C, and D. Incorrect signs will lead to incorrect results.
• Data Accuracy: The accuracy of your final answer depends on the accuracy of the given data ($\Delta H^{\circ}_{f}$ and A, B, C, D values). Use reliable sources for these values.
• Assumptions: This method assumes that the heat capacities are well-represented by the given polynomial expression over the temperature range of 25°C to 300°C. If the temperature range is much larger, this assumption may not be valid.

By following these steps carefully, you can calculate the standard heat of reaction at different temperatures. Good luck, and happy calculating!