Question

Given the following data:

Answer 1

The enthalpy of the given reaction is $\boxed{{\text{155}}{\text{.65}}\;{\text{kJ}}}$ .

Further Explanation:

This problem is based upon Hess’s Law. Hess’s law is utilized for calculating the enthalpy change of a reaction that can be obtained simply by summation of two or more reactions. In accordance with the Hess’s law, $\Delta H$ of an overall reaction is obtained by adding the enthalpy change for each individual step reaction involved to obtain the overall reaction.

$\boxed{\Delta\text{H}_{\text{overall rxn}}=\Delta \text{H}_{1}+\Delta \text{H}_{2}+......+\Delta \text{H}_{n}}$

Enthalpy is defined as state function and therefore its value depends upon the initial and final state of system but not upon the path. This is the reason that the overall reaction can be simply obtained by adding or subtracting the enthalpy change of the individual steps utilized to get the final reaction.

Step 1: The enthalpy change of the following reaction is  $\Delta {H_1}$.

${{\text{N}}_2}\left(g\right)+{{\text{O}}_2}\left(g\right)\to2{\text{NO}}\left(g\right)$        ......(1)

The value of $\Delta {H_1}$ is $180.7{\text{ kJ}}$.

Step 2: The enthalpy change of the following reaction is $\Delta {H_2}$ $2{\text{NO}}\left(g\right)+{{\text{O}}_{\text{2}}}\left(g\right)\to2{\text{N}}{{\text{O}}_{\text{2}}}\left(g\right)$.                ......(2)

Step 3: The enthalpy change of the following reaction is $\Delta {H_3}$ .

$2{{\text{N}}_2}{\text{O}}\left(g\right)\to2{{\text{N}}_{\text{2}}}\left(g\right)+{{\text{O}}_2}\left(g\right)$            ......(3)

Step 4: Reverse and divide the equation (2) by 2.

${\text{N}}{{\text{O}}_{\text{2}}}\left( g\right)\to{\text{NO}}\left(g\right)+\frac{1}{2}{{\text{O}}_{\text{2}}}\left(g\right)$                                          ......(4)

Step 5: The enthalpy change for the reaction (4) is calculated as,

$\begin{aligned}\Delta{H_4}&=-\frac{{\Delta {H_2}}}{2}\\&=-\left({\frac{{ - 113.1\;{\text{kJ}}}}{2}}\right)\\&=56.55\;{\text{kJ}}\\\end{aligned}$

Step 6: Divide the equation (3) by 2.

${{\text{N}}_2}{\text{O}}\left(g\right)\to{{\text{N}}_{\text{2}}}\left(g\right)+\frac{1}{2}{{\text{O}}_2}\left(g\right)$               ......(5)

Step 7: The enthalpy change for the reaction (5) is calculated as,

$\begin{aligned}\Delta {H_5}&=\frac{{\Delta {H_3}}}{2}\\&=\frac{{-163.2\;{\text{kJ}}}}{2}\\&=- 81.6\;{\text{kJ}}\\\end{aligned}$

Step 8: Add equation (1), (4), and (5) to get the final equation

${{\text{N}}_2}{\text{O}}\left(g\right)+{\text{N}}{{\text{O}}_2}\left(g\right)\to 3{\text{NO}}\left(g\right)$             ......(6)

Step 9: The expression to calculate is as follows:

$\Delta{H_6}=\Delta {H_1}+\Delta{H_4}+\Delta{H_5}$    .......(7)

Substitute $180.7{\text{ kJ}}$ for $\Delta {H_1}$ , $56.55\;{\text{kJ}}$ for $\Delta {H_4}$ and $- 81.6\;{\text{kJ}}$ for $\Delta {H_3}$ in the equation (7).

$\begin{aligned}\Delta{H_6}&=\left({180.7{\text{kJ}}}\right)+\left({56.55\;{\text{kJ}}}\right)+\left({-81.6\;{\text{kJ}}}\right)\\&=155.65\;{\text{kJ}}\\\end{aligned}$

Hence, the enthalpy of the given reaction (6) is ${\mathbf{155}}{\mathbf{.65}}\;{\mathbf{kJ}}$.

Learn more:

1. Oxidation and reduction reaction brainly.com/question/2973661

2. Calculation of moles of HCl: brainly.com/question/5950133

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Thermodynamics

Keywords: Hess’s Law, enthalpy, NO, NO2, N2O, 180.7 kj, 113.1 kj, 163.2 kj , overall reaction, adding, state function, initial state, final state and 155.65 kj

Answer 2

Answer is: enthalpy is 155.65 kJ.

Reaction 1: N₂(g) + O₂(g) → 2NO(g); ΔrH₁ = 180.7 kJ.

Reaction 2: 2NO(g) + O₂(g) → 2NO₂(g); ΔrH₂ = -113.1 kJ.

Reaction 3: 2N₂O → 2N₂(g) + O₂(g); ∆rH₃ = -163.2 kJ.

Reaction 4: N₂O(g) + NO₂(g) → 3NO(g); ΔrH₄ = ?.

Using Hess's law reaction number 4 is sum of reaction number 1 and half of reaction number 3 minus half of the reaction number 2.

ΔrH₄ = ΔrH₁ + 1/2ΔrH₃ - 1/2ΔrH₂.

ΔrH₄ = 180.7 kJ + 1/2 · (-163.2 kJ) - 1/2 · (-113.1 kJ).

ΔrH₄ = 155.65 kJ.