When 9.41 g of phenol, C6H5OH, are burned in a bomb calorimeter at 25 oC, 305.1 kJ of heat are given off. How to calculate the standard enthalpy of combustion, in kJ/mol, of phenol?

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When 9.41 g of phenol, C6H5OH, are burned in a bomb calorimeter at 25 oC, 305.1 kJ of heat are given off. How to calculate the standard enthalpy of combustion, in kJ/mol, of phenol?

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Answer 1 · 2015-02-28T22:11:42

The standard enthalpy of combustion of phenol is $-3051 kJ/mol$ $"-3051 kJ/mol"$ .

You can solve this problem without the balanced chemical equation and without standard enthalpies of formation, all you have to do is use the heat absorbed by the water.

So, you're burning phenol, $C_{6} H_{5} O H$ $C_6H_5OH$ , in a bomb calorimeter at $25^{\circ} C$ $"25"^@"C"$ . Notice that you have a positive value for the heat given off in the reaction, which means that this represents the heat absorbed by the water.

The actual heat given off by the combustion of phenol will be equal to this value but have a negative sign - heat released, not absorbed

$Δ H_{phenol} = - q_{water}$ $DeltaH_("phenol") = -q_("water")$

$Δ H_{phenol} = - 305.1 kJ$ $DeltaH_("phenol") = -"305.1 kJ"$

This is how much heat is given off in the combustion of $9.41 g$ $"9.41 g"$ of phenol; however, what you need to determine is the standard enthalpy of combustion, which is defined as the enthalpy change when 1 mole of a substance is burned completely under standard conditions.

You need to determine exactly how many moles of phenol you've burned

$9.41 g phenol \cdot \frac{1 mole}{94.1 g} = 0.100 moles phenol$ $"9.41 g phenol" * "1 mole"/"94.1 g" = "0.100 moles phenol"$

This means that you give off -305.1 kJ when 0.100 moles are burned; therefore, when 1 mole is burned you'll give off 10 times as much heat

$1.000 moles \cdot \frac{-305.1 kJ}{0.100 moles} = -3051 kJ$ $"1.000 moles" * "-305.1 kJ"/"0.100 moles" = "-3051 kJ"$

SInce this value represents the heat given off per mole, you can write it as

$\frac{-3051 kJ}{1 mole} = -3051 kJ/mol$ $"-3051 kJ"/"1 mole" = "-3051 kJ/mol"$

SIDE NOTE Rounded to 3 sig figs, the answer wil be -3050 kJ/mol.
I assume the pressure to be at 1 atm, since standard state implies a pressure of 1 atm and a temperature of $25^{\circ} C$ $25^@"C"$ . Moreover, all the species involved in the reaction must be in their standard states, so your balanced chemical equation will look like this

$C_{6} H_{5} O H_{(s)} + 7 O_{2 (g)} \to 6 C O_{2 (g)} + 3 H_{2} O_{(l)}$ $C_6H_5OH_((s)) + 7O_(2(g)) ->6CO_(2(g)) + 3H_2O_((l))$

Here is a video which discusses how to calculate the enthalpy change when 0.13g of butane is burned.

Video from: Noel Pauller

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When 9.41 g of phenol, C6H5OH, are burned in a bomb calorimeter at 25 oC, 305.1 kJ of heat are given off. How to calculate the standard enthalpy of combustion, in kJ/mol, of phenol?

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