1. Tardigrade
  2. Question
  3. Chemistry
  4. The value of enthalpy change (ΔH) for the reaction C2H5OH(I) + 3O2(g) → 2CO2(g) + 3H2O(I) 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 ($\Delta$H) for the reaction
$C_{2}H_{5}OH_{\left(I\right)} + 3O_{2\left(g\right)} \to 2CO_{2\left(g\right)} + 3H_{2}O_{\left(I\right)}$
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 :

AIEEEAIEEE 2011Thermodynamics

Solution:

$C_{2}H_{5}OH\left(\ell\right) + 3O_{2\left(g\right)} \to 2CO_{2\left(g\right)} + 3H_{2}O\left(\ell\right)$
$\Delta n_{g} = 2-3 = -1$
$\Delta U = \Delta H - \Delta n_{g} RT$
$= -1366.5-\left(-1\right)\times\frac{8.314}{10^{3}}\times300$
$= - 1366.5 + 0.8314 × 3 = - 1364 KJ$