C ₂ H ₅ OH (l) + 3 O ₂(g) 2 CO ₂(g) + 3 H ₂ O (l) The value of enthalpy change ( H ) for above reaction at 27 ,^ C is -1366.5 kJ mol ⁻¹ . Then value of internal energy change for the same reaction at this temperature will be

Step-by-step solution

The given chemical equation is: C ₂ H ₅ OH (l) + 3 O ₂(g) 2 CO ₂(g) + 3 H ₂ O (l) The change in the number of moles of gaseous species, n_g , is calculated as: n_g = n_ p(g) - n_ r(g) = 2 - 3 = -1 The relationship between enthalpy change ( H ) and internal energy change ( U ) is given by: H = U + n_g RT Given: H = -1366.5 kJ mol ⁻¹ R = 8.314 10⁻³ kJ K ⁻¹ mol ⁻¹ T = 27^ C = 300 K Substituting the values into the equation: -1366.5 = U + (-1) (8.314 10⁻³) 300 -1366.5 = U - 2.4942 U = -1366.5 + 2.4942 = -1364.0058 kJ m