Question

given the thermochemical equations

x₂ + 3y₂ → 2xy₃ δh₁ = -390 kj

x₂ + 2z₂ → 2xz₂ δh₂ = -110 kj

2y₂ + z₂ → 2y₂z δh₃ = -280 kj

calculate the change in enthalpy for the reaction.

4xy₃ + 7z₂ → 6y₂z + 4xz₂

δh = kj

Explanation:

Step1: Reverse the first equation

Reverse \(X_{2}+3Y_{2}\to2XY_{3}\) (\(\Delta H_{1}=- 390\ \text{kJ}\)) to get \(2XY_{3}\to X_{2}+3Y_{2}\), and \(\Delta H_{1}^{'} = 390\ \text{kJ}\). Multiply this reversed equation by \(2\): \(4XY_{3}\to2X_{2}+6Y_{2}\), \(\Delta H_{a}=2\times390\ \text{kJ} = 780\ \text{kJ}\)

Step2: Manipulate the second equation

Multiply \(X_{2}+2Z_{2}\to2XZ_{2}\) (\(\Delta H_{2}=-110\ \text{kJ}\)) by \(2\): \(2X_{2}+4Z_{2}\to4XZ_{2}\), \(\Delta H_{b}=2\times(-110)\ \text{kJ}=-220\ \text{kJ}\)

Step3: Manipulate the third equation

Multiply \(2Y_{2}+Z_{2}\to2Y_{2}Z\) (\(\Delta H_{3}=-280\ \text{kJ}\)) by \(3\): \(6Y_{2}+3Z_{2}\to6Y_{2}Z\), \(\Delta H_{c}=3\times(-280)\ \text{kJ}=-840\ \text{kJ}\)

Step4: Apply Hess's law

Add the three manipulated equations (\(4XY_{3}\to2X_{2}+6Y_{2}\), \(2X_{2}+4Z_{2}\to4XZ_{2}\), \(6Y_{2}+3Z_{2}\to6Y_{2}Z\)) together:
\(4XY_{3}+(2X_{2}- 2X_{2})+(6Y_{2}-6Y_{2})+(4Z_{2}+3Z_{2})\to6Y_{2}Z + 4XZ_{2}\), which simplifies to \(4XY_{3}+7Z_{2}\to6Y_{2}Z + 4XZ_{2}\)
By Hess's law \(\Delta H=\Delta H_{a}+\Delta H_{b}+\Delta H_{c}\)
\(\Delta H=780-220 - 840\ \text{kJ}\)

Answer:

\(-280\ \text{kJ}\)