Question

For the complete combustion of ethanol, $C_2H_5OH_{(l)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)},$ the amount of heat produced as measured in bomb calorimeter, is $1364.47 \,kJ \,moI^{-1}$ at $25^{\circ}$C. Assuming ideality the enthalpy of combustion, $\Delta_ CH$, for the reaction will be $(R = 8.314\, J \,K^{-1}$ $mol^{-1})$

• A. $- 1366.95 \,kJ \,moI^{-1}$
• B. $- 1361.95 \, kJ \, moI^{-1}$
• C. $- 1460.50 \, kJ \,moI^{-1}$
• D. $- 1350.50\, kJ \, moI^{-1}$

Answer 1

The Correct Option is A

The problem involves calculating the enthalpy of combustion (\(\Delta_C H\)) from the given heat measured in a bomb calorimeter. The reaction is given as:

C_2H_5OH_{(l)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}

Here, the heat produced in a bomb calorimeter is given as 1364.47 \,kJ \,mol^{-1}. This value represents the internal energy change, \(\Delta U\), for the combustion process. To find the enthalpy change, \(\Delta_C H\), we need to relate \(\Delta U\) with \(\Delta H\) via the formula:

\Delta H = \Delta U + \Delta n_gRT

Where:

• \(\Delta n_g\) is the change in moles of gas during the reaction.
• \(R\) is the universal gas constant \(8.314 \, J \, K^{-1} \, mol^{-1}\).
• \(T\) is the temperature in Kelvin (25°C = 298K).

For the reaction, calculate \(\Delta n_g\):

• Moles of gas before reaction = 3 moles of \(O_2\).
• Moles of gas after reaction = 2 moles of \(CO_2\).

Thus, \(\Delta n_g = 2 - 3 = -1\).

Convert \(R\) from Joules to kJ:

R = 8.314 \, J \, K^{-1} \, mol^{-1} = 0.008314 \, kJ \, K^{-1} \, mol^{-1}

Substitute the values into the enthalpy equation:

\Delta H = 1364.47 \, kJ \, mol^{-1} + (-1)(0.008314 \, kJ \, K^{-1} \, mol^{-1})(298 \, K)

Calculate the second term:

= -(0.008314 \times 298)

= -2.478 \, kJ \, mol^{-1}

Therefore,

\Delta H = 1364.47 - 2.478 = 1366.948 \, kJ \, mol^{-1}

Rounding to two decimal places, \(\Delta H\) is approximately \(-1366.95 \, kJ \, mol^{-1}\).

The correct answer is - 1366.95 \,kJ \,mol^{-1}.