Question

calculate the standard enthalpy of formation, $delta h_{f}^{circ}$, of liquid $ce{n_{2}h_{4}}$ using the information below. balanced equation: $2ce{nh_{3}(g)} + \frac{1}{2}ce{o_{2}(g)}
ightarrow ce{n_{2}h_{4}(l)} + ce{h_{2}o(l)}$ enthalpy of combustion: $delta h_{\text{combustion}(ce{nh_{3}})} = -71.5 \text{kj/mol} ce{nh_{3}}$ enthalpies of formation: | compound | $delta h_{f}^{circ}$ (kj/mol) | | --- | --- | | $ce{nh_{3}(g)}$ | -46.0 | | $ce{h_{2}o(l)}$ | -285.8 | | $ce{n_{2}h_{4}(l)}$ |? | enter your answer using either a + or - and the magnitude. hint: be sure to take the number of moles of $ce{nh_{3}}$ into account for the enthalpy of combustion.

Explanation:

Step1: Recall the formula for enthalpy change of a reaction

The enthalpy change of a reaction ($\Delta H_{rxn}$) is given by the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants, i.e., $\Delta H_{rxn} = \sum \Delta H_f^{\circ}(\text{products}) - \sum \Delta H_f^{\circ}(\text{reactants})$. Also, the enthalpy of combustion of $\ce{NH3}$ is related to the reaction. For the given reaction: $2\ce{NH3}(g) + \frac{1}{2}\ce{O2}(g)
ightarrow \ce{N2H4}(l) + \ce{H2O}(l)$, the enthalpy change of this reaction can also be calculated using the enthalpy of combustion of $\ce{NH3}$. The enthalpy of combustion of 1 mole of $\ce{NH3}$ is -71.5 kJ/mol, so for 2 moles, it will be $2 \times (-71.5)$ kJ.

Step2: Calculate $\Delta H_{rxn}$ from combustion data

$\Delta H_{rxn} = 2 \times \Delta H_{\text{combustion}}(\ce{NH3}) = 2 \times (-71.5) = -143.0$ kJ.

Step3: Apply the enthalpy of formation formula to the reaction

$\Delta H_{rxn} = [\Delta H_f^{\circ}(\ce{N2H4}(l)) + \Delta H_f^{\circ}(\ce{H2O}(l))] - [2 \times \Delta H_f^{\circ}(\ce{NH3}(g)) + \frac{1}{2} \times \Delta H_f^{\circ}(\ce{O2}(g))]$. Since the enthalpy of formation of an element in its standard state (like $\ce{O2}(g)$) is 0, $\Delta H_f^{\circ}(\ce{O2}(g)) = 0$. Substituting the known values:
$-143.0 = [\Delta H_f^{\circ}(\ce{N2H4}(l)) + (-285.8)] - [2 \times (-46.0) + 0]$

Step4: Solve for $\Delta H_f^{\circ}(\ce{N2H4}(l))$

First, simplify the right-hand side:
$[\Delta H_f^{\circ}(\ce{N2H4}(l)) - 285.8] - [-92.0] = \Delta H_f^{\circ}(\ce{N2H4}(l)) - 285.8 + 92.0 = \Delta H_f^{\circ}(\ce{N2H4}(l)) - 193.8$

Now, set equal to $\Delta H_{rxn}$:
$\Delta H_f^{\circ}(\ce{N2H4}(l)) - 193.8 = -143.0$

Add 193.8 to both sides:
$\Delta H_f^{\circ}(\ce{N2H4}(l)) = -143.0 + 193.8 = 50.8$ kJ/mol. Wait, that can't be right. Wait, let's check the signs again. Wait, maybe I messed up the direction. Wait, no, let's re-express the formula correctly.

Wait, the correct formula is $\Delta H_{rxn} = \sum \Delta H_f^{\circ}(\text{products}) - \sum \Delta H_f^{\circ}(\text{reactants})$. So products are $\ce{N2H4}(l)$ and $\ce{H2O}(l)$, reactants are $2\ce{NH3}(g)$ and $\frac{1}{2}\ce{O2}(g)$. So:

$\Delta H_{rxn} = [\Delta H_f^{\circ}(\ce{N2H4}) + \Delta H_f^{\circ}(\ce{H2O})] - [2\Delta H_f^{\circ}(\ce{NH3}) + \frac{1}{2}\Delta H_f^{\circ}(\ce{O2})]$

We know $\Delta H_{rxn} = 2 \times \Delta H_{\text{combustion}}(\ce{NH3}) = 2 \times (-71.5) = -143$ kJ (since combustion is exothermic, the reaction as written has $\Delta H$ equal to the combustion enthalpy for 2 moles).

Substitute the known $\Delta H_f^{\circ}$ values:

$-143 = [\Delta H_f^{\circ}(\ce{N2H4}) + (-285.8)] - [2 \times (-46.0) + 0]$ (since $\Delta H_f^{\circ}(\ce{O2}) = 0$)

Simplify the right-hand side:

$[\Delta H_f^{\circ}(\ce{N2H4}) - 285.8] - [-92] = \Delta H_f^{\circ}(\ce{N2H4}) - 285.8 + 92 = \Delta H_f^{\circ}(\ce{N2H4}) - 193.8$

Now, solve for $\Delta H_f^{\circ}(\ce{N2H4})$:

$\Delta H_f^{\circ}(\ce{N2H4}) - 193.8 = -143$

$\Delta H_f^{\circ}(\ce{N2H4}) = -143 + 193.8 = 50.8$? Wait, but let's check again. Wait, maybe the reaction is the combustion of $\ce{NH3}$, but the given reaction is not the combustion reaction. Wait, the combustion of $\ce{NH3}$ would be $\ce{NH3} + \frac{3}{4}\ce{O2}
ightarrow \frac{1}{2}\ce{N2} + \frac{3}{2}\ce{H2O}$, but the given reaction is different. Wait, maybe I made a mistake in relating the reaction to the combustion. Let's re-express the given reaction.

Wait, the enthalpy of combustion of $\ce{NH3}$ is the enthalpy change when 1 mole of $\ce{NH3}$ is combusted. But the given reaction is $2\ce{NH3} + \frac{1}{2}\ce{O2}
ightarrow \ce{N2H4} + \ce{H2O}$. So let's calculate $\Delta H_{rxn}$ using the combustion data. The combustion of $\ce{NH3}$: $\ce{NH3}(g) + \frac{3}{4}\ce{O2}(g)
ightarrow \frac{1}{2}\ce{N2}(g) + \frac{3}{2}\ce{H2O}(l)$, $\Delta H = -71.5$ kJ/mol. But our reaction is different. Wait, maybe the problem is that the given reaction's enthalpy change is equal to the enthalpy of combustion for 2 moles of $\ce{NH3}$? Wait, no, let's re-express the given reaction.

Alternative approach: Let's denote $\Delta H_f^{\circ}(\ce{N2H4}) = x$. Then,

$\Delta H_{rxn} = (x + (-285.8)) - (2 \times (-46.0) + 0)$ (since $\ce{O2}$ is element, $\Delta H_f = 0$)

$\Delta H_{rxn} = x - 285.8 + 92 = x - 193.8$

But also, the enthalpy change of the reaction is equal to the enthalpy of combustion for 2 moles of $\ce{NH3}$, which is $2 \times (-71.5) = -143$ kJ. So:

$x - 193.8 = -143$

$x = -143 + 193.8 = 50.8$? But that seems positive, but let's check with another method. Wait, maybe the reaction is not the combustion, but the enthalpy of combustion is given, so maybe the given reaction's $\Delta H$ is equal to the enthalpy of combustion for 2 moles. Wait, perhaps I messed up the sign. Wait, combustion is exothermic, so $\Delta H$ is negative. The given reaction: let's see the reactants and products. $\ce{NH3}$ is reacting, $\ce{N2H4}$ and $\ce{H2O}$ are products. Let's check the oxidation states. Nitrogen in $\ce{NH3}$ is -3, in $\ce{N2H4}$ is -2, so it's being oxidized, and oxygen is being reduced (from 0 in $\ce{O2}$ to -2 in $\ce{H2O}$). So the reaction is a redox reaction. But the enthalpy of combustion of $\ce{NH3}$ is the enthalpy change when $\ce{NH3}$ is burned in oxygen to form stable products (like $\ce{N2}$ and $\ce{H2O}$), but in our reaction, the product is $\ce{N2H4}$ (not $\ce{N2}$), so maybe the given enthalpy of combustion is for a different reaction, but the problem states to use that. So proceeding with the calculation:

Wait, maybe I made a mistake in the formula. Let's write the formula again:

$\Delta H_{rxn} = \sum n \Delta H_f^{\circ}(\text{products}) - \sum n \Delta H_f^{\circ}(\text{reactants})$

For the reaction: $2\ce{NH3}(g) + \frac{1}{2}\ce{O2}(g)
ightarrow \ce{N2H4}(l) + \ce{H2O}(l)$

Products: $\ce{N2H4}(l)$ (1 mol) and $\ce{H2O}(l)$ (1 mol)

Reactants: $\ce{NH3}(g)$ (2 mol) and $\ce{O2}(g)$ (0.5 mol)

So,

$\Delta H_{rxn} = [\Delta H_f^{\circ}(\ce{N2H4}) + \Delta H_f^{\circ}(\ce{H2O})] - [2\Delta H_f^{\circ}(\ce{NH3}) + 0.5\Delta H_f^{\circ}(\ce{O2})]$

Since $\Delta H_f^{\circ}(\ce{O2}) = 0$,

$\Delta H_{rxn} = \Delta H_f^{\circ}(\ce{N2H4}) + \Delta H_f^{\circ}(\ce{H2O}) - 2\Delta H_f^{\circ}(\ce{NH3})$

We also know that the enthalpy change of this reaction is equal to the enthalpy of combustion of 2 moles of $\ce{NH3}$, which is $2 \times (-71.5) = -143$ kJ (because combustion of 1 mole is -71.5, so 2 moles is 2*(-71.5)).

Substituting the known $\Delta H_f^{\circ}$ values:

$-143 = \Delta H_f^{\circ}(\ce{N2H4}) + (-285.8) - [2*(-46.0) + 0]$

Simplify the right-hand side:

$\Delta H_f^{\circ}(\ce{N2H4}) - 285.8 + 92 = \Delta H_f^{\circ}(\ce{N2H4}) - 193.8$

Now, solve for $\Delta H_f^{\circ}(\ce{N2H4})$:

$\Delta H_f^{\circ}(\ce{N2H4}) - 193.8 = -143$

$\Delta H_f^{\circ}(\ce{N2H4}) = -143 + 193.8 = 50.8$ kJ/mol. Wait, but let's check with actual values. The standard enthalpy of formation of $\ce{N2H4}(l)$ is actually about +50.6 kJ/mol, so this matches. So the calculation is correct.

Answer:

+50.8