Diketahui reaksi: $$ \mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{HCl}(g) $$ Jika energi disosiasi ikatan: $$ \mathrm{C}-\mathrm{H}=413 \mathrm{~kJ} / \mathrm{mol} $$ $$ \mathrm{C}-\mathrm{Cl}=328 \mathrm{~kJ} / \mathrm{mol} $$ $$ \mathrm{Cl}-\mathrm{Cl}=242 \mathrm{~kJ} / \mathrm{mol} $$ $$ \mathrm{H}-\mathrm{Cl}=431 \mathrm{~kJ} / \mathrm{mol} $$ maka tentukan $$ \Delta \mathrm{H} $$ reaksi tersebut!

Question

UpStudy AI · Accepted Answer

Alright, I need to determine the enthalpy change ($$ \Delta H $$) for the given chemical reaction:

$$ \mathrm{CH}_{4}(g) + \mathrm{Cl}_{2}(g) \rightarrow \mathrm{CH}_{3}\mathrm{Cl}(g) + \mathrm{HCl}(g) $$

To calculate $$ \Delta H $$, I'll use the standard enthalpy of formation ($$ \Delta H_f^\circ $$) values of the reactants and products. The formula for $$ \Delta H $$ is:

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

First, I need to find the $$ \Delta H_f^\circ $$ values for each species involved in the reaction. However, the problem provides bond dissociation energies instead of standard enthalpies of formation. Bond dissociation energies are the energy required to break a specific bond in a molecule. To relate bond dissociation energies to $$ \Delta H_f^\circ $$, I'll need to consider the bonds broken and formed during the reaction.

Let's identify the bonds broken and formed:

**Reactants:**
- $$ \mathrm{CH}_4(g) $$: This molecule has four $$ \mathrm{C}-\mathrm{H} $$ bonds.
- $$ \mathrm{Cl}_2(g) $$: This molecule has one $$ \mathrm{Cl}-\mathrm{Cl} $$ bond.

**Products:**
- $$ \mathrm{CH}_3\mathrm{Cl}(g) $$: This molecule has three $$ \mathrm{C}-\mathrm{H} $$ bonds and one $$ \mathrm{C}-\mathrm{Cl} $$ bond.
- $$ \mathrm{HCl}(g) $$: This molecule has one $$ \mathrm{H}-\mathrm{Cl} $$ bond.

Now, let's calculate the total bond dissociation energy for the reactants and products.

**Reactants:**
- Breaking four $$ \mathrm{C}-\mathrm{H} $$ bonds: $$ 4 \times 413 \, \mathrm{kJ/mol} = 1652 \, \mathrm{kJ/mol} $$
- Breaking one $$ \mathrm{Cl}-\mathrm{Cl} $$ bond: $$ 1 \times 242 \, \mathrm{kJ/mol} = 242 \, \mathrm{kJ/mol} $$
- **Total for reactants:** $$ 1652 + 242 = 1894 \, \mathrm{kJ/mol} $$

**Products:**
- Forming three $$ \mathrm{C}-\mathrm{H} $$ bonds: $$ 3 \times 413 \, \mathrm{kJ/mol} = 1239 \, \mathrm{kJ/mol} $$
- Forming one $$ \mathrm{C}-\mathrm{Cl} $$ bond: $$ 1 \times 328 \, \mathrm{kJ/mol} = 328 \, \mathrm{kJ/mol} $$
- Forming one $$ \mathrm{H}-\mathrm{Cl} $$ bond: $$ 1 \times 431 \, \mathrm{kJ/mol} = 431 \, \mathrm{kJ/mol} $$
- **Total for products:** $$ 1239 + 328 + 431 = 1998 \, \mathrm{kJ/mol} $$

Now, to find $$ \Delta H $$, subtract the total bond dissociation energy of the products from that of the reactants:

$$
\Delta H = \text{Total for reactants} - \text{Total for products} = 1894 \, \mathrm{kJ/mol} - 1998 \, \mathrm{kJ/mol} = -104 \, \mathrm{kJ/mol}
$$

Since the result is negative, the reaction is exothermic, releasing 104 kJ of energy per mole of reaction.

The enthalpy change ($$ \Delta H $$) for the reaction is -104 kJ/mol.

Real Tutor Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Chemistry Questions