Question

When $ 9.41g $ of phenol, $ {C_6}{H_5}OH $ , are burned in a bomb calorimeter at $ {25^ \circ }C $ , $ 305.1\;kJ $ heat is given off. How to calculate the standard enthalpy of combustion, in $ kJmo{l^{ - 1}} $ , of phenol?

Answer 1

Hint: In this question, we will first find the enthalpy of phenol when $ 305.1\;kJ $ of heat is given off. After that we will find the number of moles of phenol we will burn. Then will find the standard enthalpy of phenol for which we will find the enthalpy of phenol of one mole.

Complete Step By Step Answer:
In this question, we can see that the heat given off in the reaction has a positive value, this indicates that the heat is absorbed by the water. The actual value of heat given off during combustion of phenol will be equal to this value but have negative signs as heat is released but not absorbed.
 $ \Delta {H_{phenol}} = - {q_{water}} $
 $ \Delta {H_{phenol}} = - 305.1\;kJ $
This much heat is given off in the combustion of $ 9.41g $ of phenol. Moreover, we have to find the standard enthalpy of combustion, which is the change in the enthalpy when one mole of the substance is burned completely under standard conditions. For that we have to find the number of moles of phenol we have burned.
 $ 9.41g\;phenol \times \dfrac{{1\;mole}}{{94.1g}} $
 $ = 0.100\;moles\;phenol $
This indicates that we get $ - 305.1\;kJ $ when we burn $ 0.100\,moles $ . Therefore, when we will burn $ 1 $ mole we will get:
 $ \Delta {H_{phenol}} = - \dfrac{{305.1\;kJ}}{{0.100\;moles}} $
 $ \Rightarrow \Delta {H_{phenol}} = - 3051kJ $
Thus, the standard enthalpy of combustion of phenol is $ - 3051\;kJ $ .

Note:
Standard enthalpy of combustion $ \left( {\Delta H_c^ \circ } \right) $ is the change in the enthalpy when one mole of the substance burns under standard state conditions. It is also known as “heat of combustion.” Many substances with large enthalpies of combustion are used as fuels like carbon, hydrogen etc.