Calculate H for a reaction, 2X_ (g) + Y_ (g) 2Z_ (g) at 298 K if U = -10.5 kJ and R = 8.314 J K ⁻¹ mol ⁻¹ .

Step-by-step solution

For the given reaction 2X_ (g) + Y_ (g) 2Z_ (g) , the change in the number of moles of gaseous species is: n_g = moles of gaseous products - moles of gaseous reactants n_g = 2 - (2 + 1) = -1 The relationship between enthalpy change ( H ) and internal energy change ( U ) is given by: H = U + n_g RT Substituting the given values: H = -10.5 kJ + (-1) (8.314 10⁻³ kJ K ⁻¹ mol ⁻¹) 298 K H = -10.5 kJ - 2.477 kJ H = -12.977 kJ -12.98 kJ Answer: -12.98 kJ