Question

If 3 moles of a compound use 12 J of energy in a reaction, what is the Hreaction in kJ/mol

Answer 1

Answer:

$\Delta _RH=4x10^{-3}\frac{kJ}{mol}$

Explanation:

Hello,

In this case, the molar enthalpy of reaction is obtained by dividing the involved energy by the reacting moles:

$\Delta _RH=\frac{12J}{3mol} =4\frac{J}{mol}$

Thus, it is important to notice that the compound "uses" the energy, it means that it absorbs the energy, for that reason the sign is positive. Moreover, computing the result in kJ/mol we finally obtain:

$\Delta _RH=4\frac{J}{mol}*\frac{1kJ}{1000J} =4x10^{-3}\frac{kJ}{mol}$

Best regards.