Question

calculate the enthalpy for this reaction.
mg + \frac{1}{2}o_2 \
ightarrow mgo
use the following thermochemical equations.

1. h_2 + \frac{1}{2}o_2 \

ightarrow h_2o \quad \delta h_1 = -286 kj

1. mg + 2hcl \

ightarrow mgcl_2 + h_2 \quad \delta h_2 = -462 kj

1. mgo + 2hcl \

ightarrow mgcl_2 + h_2o \quad \delta h_3 = -105 kj
\delta h_{rxn} = ? kj
enter either a + or - sign and the magnitude.
use significant figures.

Explanation:

Step1: Identify target and given reactions

Target: \( \text{Mg} + \frac{1}{2}\text{O}_2
ightarrow \text{MgO} \)
Given:

1. \( \text{H}_2 + \frac{1}{2}\text{O}_2

ightarrow \text{H}_2\text{O} \quad \Delta H_1 = -286 \, \text{kJ} \)

1. \( \text{Mg} + 2\text{HCl}

ightarrow \text{MgCl}_2 + \text{H}_2 \quad \Delta H_2 = -462 \, \text{kJ} \)

1. \( \text{MgO} + 2\text{HCl}

ightarrow \text{MgCl}_2 + \text{H}_2\text{O} \quad \Delta H_3 = -105 \, \text{kJ} \)

Step2: Manipulate equations to sum to target

• Reverse equation 3: \( \text{MgCl}_2 + \text{H}_2\text{O}

ightarrow \text{MgO} + 2\text{HCl} \quad \Delta H_3' = +105 \, \text{kJ} \) (reverse, so sign flips)

• Add equations 1, 2, and reversed 3:
• Eq1: \( \text{H}_2 + \frac{1}{2}\text{O}_2

ightarrow \text{H}_2\text{O} \)

• Eq2: \( \text{Mg} + 2\text{HCl}

ightarrow \text{MgCl}_2 + \text{H}_2 \)

• Reversed Eq3: \( \text{MgCl}_2 + \text{H}_2\text{O}

ightarrow \text{MgO} + 2\text{HCl} \)

• Sum: \( \text{Mg} + \frac{1}{2}\text{O}_2

ightarrow \text{MgO} \) (HCl, MgCl₂, H₂, H₂O cancel)

Step3: Calculate total enthalpy

\( \Delta H_{\text{rxn}} = \Delta H_1 + \Delta H_2 + \Delta H_3' \)
\( \Delta H_{\text{rxn}} = -286 + (-462) + 105 \)
\( \Delta H_{\text{rxn}} = -286 - 462 + 105 \)
\( \Delta H_{\text{rxn}} = -643 \, \text{kJ} \)

Answer:

\(-643\)