Ethyne $$ \left(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}), \Delta H_{\mathrm{f}}=226.77 \mathrm{~kJ} / \mathrm{mol}\right) $$ undergoes complete combustion in the presence of oxygen to produce carbon dioxide $$ \left(\mathrm{CO}_{2}(\mathrm{~g}), \Delta H_{\mathrm{f}}=-393.5 \mathrm{~kJ} / \mathrm{mol}\right) $$ and water $$ \left(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}), \Delta H_{\mathrm{f}}=-241.82 \mathrm{~kJ} / \mathrm{mol}\right) $$ according to the equation below. $$ 2 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ What is the enthalpy of combustion (per mole) of $$ \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) $$ ? Use $$ \Delta H_{\text {rnn }}=\sum\left(\Delta H_{\text {f,prowucts }}\right)-\sum\left(\Delta H_{\text {f, reactants }}\right) $$. $$ -2511.2 \mathrm{~kJ} / \mathrm{mol} $$ $$ -1255.6 \mathrm{~kJ} / \mathrm{mol} $$ $$ -862.1 \mathrm{~kJ} / \mathrm{mol} $$ $$ -431.0 \mathrm{~kJ} / \mathrm{mol} $$

Question

UpStudy AI · Accepted Answer

To find the enthalpy of combustion of ethyne ($$ \mathrm{C}_{2} \mathrm{H}_{2} $$), we can use the given reaction and the standard enthalpies of formation ($$ \Delta H_f $$) for the reactants and products.

The balanced combustion reaction is:

$$
2 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g}) + 5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{g}) + 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})
$$

### Step 1: Calculate the total enthalpy of formation for the products

1. For $$ \mathrm{CO}_{2} $$:
   $$
   \Delta H_f = -393.5 \, \text{kJ/mol}
   $$
   Since there are 4 moles of $$ \mathrm{CO}_{2} $$:
   $$
   4 \times (-393.5) = -1574.0 \, \text{kJ}
   $$

2. For $$ \mathrm{H}_{2} \mathrm{O} $$:
   $$
   \Delta H_f = -241.82 \, \text{kJ/mol}
   $$
   Since there are 2 moles of $$ \mathrm{H}_{2} \mathrm{O} $$:
   $$
   2 \times (-241.82) = -483.64 \, \text{kJ}
   $$

3. Total enthalpy of formation for products:
   $$
   \Delta H_f (\text{products}) = -1574.0 + (-483.64) = -2057.64 \, \text{kJ}
   $$

### Step 2: Calculate the total enthalpy of formation for the reactants

1. For $$ \mathrm{C}_{2} \mathrm{H}_{2} $$:
   $$
   \Delta H_f = 226.77 \, \text{kJ/mol}
   $$
   Since there are 2 moles of $$ \mathrm{C}_{2} \mathrm{H}_{2} $$:
   $$
   2 \times 226.77 = 453.54 \, \text{kJ}
   $$

2. For $$ \mathrm{O}_{2} $$:
   The standard enthalpy of formation for $$ \mathrm{O}_{2} $$ is 0 (as it is in its elemental form).

3. Total enthalpy of formation for reactants:
   $$
   \Delta H_f (\text{reactants}) = 453.54 + 0 = 453.54 \, \text{kJ}
   $$

### Step 3: Calculate the enthalpy change for the reaction

Using the formula:
$$
\Delta H_{\text{rxn}} = \sum \Delta H_f (\text{products}) - \sum \Delta H_f (\text{reactants})
$$
$$
\Delta H_{\text{rxn}} = -2057.64 - 453.54 = -2511.18 \, \text{kJ}
$$

### Step 4: Calculate the enthalpy of combustion per mole of $$ \mathrm{C}_{2} \mathrm{H}_{2} $$

Since the reaction involves 2 moles of $$ \mathrm{C}_{2} \mathrm{H}_{2} $$, the enthalpy of combustion per mole of $$ \mathrm{C}_{2} \mathrm{H}_{2} $$ is:
$$
\Delta H_{\text{combustion}} = \frac{-2511.18}{2} = -1255.59 \, \text{kJ/mol}
$$

### Conclusion

The enthalpy of combustion of $$ \mathrm{C}_{2} \mathrm{H}_{2} $$ is approximately:
$$
\boxed{-1255.6 \, \text{kJ/mol}}
$$
The enthalpy of combustion of ethyne is approximately -1255.6 kJ/mol.

Ethyne undergoes complete combustion in the presence of oxygen to produce carbon
dioxide and water according to the equation below.

What is the enthalpy of combustion (per mole) of ?
Use .

Upstudy AI Solution

Answer

Solution

Step 1: Calculate the total enthalpy of formation for the products

Step 2: Calculate the total enthalpy of formation for the reactants

Step 3: Calculate the enthalpy change for the reaction

Step 4: Calculate the enthalpy of combustion per mole of

Conclusion

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Ethyne undergoes complete combustion in the presence of oxygen to produce carbon dioxide and water according to the equation below. What is the enthalpy of combustion (per mole) of ? Use .