1. Tardigrade
  2. Question
  3. Chemistry
  4. The value of enthalpy change (Δ H) for the reaction C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l) at 27°C is -1366.5 kJ mol-1 The value of internal energy change for the above reaction at this temperature will be

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Q. The value of enthalpy change $\left(\Delta H\right)$ for the reaction
$C_{2}H_{5}OH\left(l\right)+3O_{2}\left(g\right)\to2CO_{2}\left(g\right)+3H_{2}O\left(l\right)$ at $ 27^{\circ}C is -1366.5\, kJ\, mol^{-1}$
The value of internal energy change for the above reaction at this temperature will be

Thermodynamics

Solution:

Relation between $\Delta H$ (enthalpy change) and $\Delta E$ (internal energy change) is $\Delta H=\Delta E+ \Delta n_{g}RT$
where $\Delta n_{g} =$ (moles of gaseous products) - (moles of gaseous reactants) For the given reaction,
$\Delta n_{g} = 2 - 3 = -1$
$\Rightarrow -1366.5=\Delta E-1\times8.314\times10^{-3}\times300$
$\therefore \Delta E=-1364.0\, kJ\, mol ^{-1}$