Question

Hydrogenation of double and triple bonds is an important industrial process. Calculate (in kJ/mole) the standard enthalpy change ΔH° for the hydrogenation of ethyne (acetylene) to ethane. Just enter a number (no units).

Answer 1

Answer:

$\Delta _RH=-310.53kJ/mol$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$C_2H_2(g)+2H_2(g)\rightarrow C_2H_6(g)$

Thus, the standard enthalpies of formation are:

$\Delta _fH_{C_2H_2}=226.73kJ/mol\\\Delta _fH_{H_2}=0kJ/mol\\\Delta _fH_{C_2H_6}=-83.8 kJ/mol$

Hence, the standard enthalpy of reaction becomes:

$\Delta _RH=-83.8kJ/mol-0kJ/mol-226.73kJ/mol\\\Delta _RH=-310.53kJ/mol$

Best regards.

Answer 2

The standard enthalpy change for the hydrogenation of ethyne to ethane is -310.5 kJ/mol.

Let's consider the reaction for the hydrogenation of ethyne to ethane.

C₂H₂(g) + 2 H₂(g) ⇒ C₂H₆(g)

The standard enthalpies of formation (ΔH°f) are:

• ΔH°f(C₂H₂(g)) = 226.7 kJ/mol
• ΔH°f(H₂(g)) = 0 kJ/mol
• ΔH°f(C₂H₆(g)) = -83.8 kJ/mol

We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.

$\Delta H\° _r = \Sigma \Delta H\°f(p) \times n_p - \Sigma \Delta H\°f(r) \times n_r$

where,

• n: moles
• r: reactants
• p: products

The standard enthalpy change for the reaction is:

$\Delta H\° _r = \Sigma \Delta H\°f(C_2H_6) \times 1mol - \Sigma \Delta H\°f(C_2H_2) \times 1mol - \Sigma \Delta H\°f(H_2) \times 2mol\\\\\Delta H\° _r = (-83.8 kJ/mol) \times 1mol - (226.7kJ/mol) \times 1mol - (0kJ/mol) \times 2mol\\\\\Delta H\° _r = -310.5 kJ$

The standard enthalpy change for the hydrogenation of ethyne to ethane is -310.5 kJ/mol.

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