Question

The enthalpy of formation for C6 H6 (I) is 49.0 kJ/mol. Consider the following reaction.

Is the reaction endothermic or exothermic, and what is the enthalpy of reaction?

exothermic; mc014-3.jpgHrxn = 49.0 kJ
exothermic; mc014-4.jpgHrxn = –49.0 kJ
endothermic; mc014-5.jpgHrxn = 49.0 kJ
endothermic; mc014-6.jpgHrxn = –49.0 kJ

Answer 1

Answer: endothermic;
`H_(rxn)=49.0 kJ`

Step-by-step explanation:

The balanced chemical reaction is,

`6C(graphite, s)+3H_2(g)\rightarrow C_6H_6(l)`

The expression for enthalpy change is,

`\Delta H=\sum [n* \Delta H_f(product)]-\sum [n* \Delta H_f(reactant)]`

`\Delta H=[(n_(C_6H_6)* \Delta H_(C_6H_6))]-[(n_(H_2)* \Delta H_(H_2))+(n_(C)* \Delta H_(C))]`

where,

n = number of moles

`\Delta H_(H_2)=0` (as heat of formation of substances in their standard state is zero

`\Delta H_{C_(graphite,s)=0` (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

`\Delta H=[(1* 49.0)]-[(3* 0)+(6* 0]`

`Delta H=49.0kJ`

Therefore, the enthalpy change for this reaction is +49.0 KJ

Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and
`\Delta H` for the reaction comes out to be negative.

Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and
`\Delta H` for the reaction comes out to be positive.

As heat change is positive, the reaction is endothermic and
`Delta H=+49.0kJ`

Answer 2

Using the formula, we subtract the enthalpy of the reactants from that of the product. The reactants have an enthalpy of 0. Thus:
ΔH = 49 - 0
ΔH = 49 kJ/mol

This means that the reaction is endothermic, with an enthalpy of 49 kJ/mol.