Calculate the work done in the oxidation of one mole HCl _ ( g ) at 27^ C , according to reaction. 4 HCl _ ( g ) + O _ 2_ ( g ) 2 Cl _ 2_ ( g ) +2 H ₂ O _ ( g ) ( R =8.314 ~J ~K ⁻¹ ~mol ⁻¹ )

Step-by-step solution

The work done during gaseous reactions under constant temperature and pressure is given by W = - n_g RT , where n_g is the change in the number of moles of gas. For 4 HCl _ ( g ) + O _ 2_ ( g ) 2 Cl _ 2_ ( g ) + 2 H ₂ O _ ( g ) , the gaseous moles are: reactants = 5, products = 4, so n_g = -1 . Per mole of HCl _ ( g ) , n_g = -1/4 = -0.25 . At T = 300 K and R = 8.314 J K⁻¹ mol⁻¹ , W = -(-0.25) 8.314 300 = 623.55 J . This value corresponds to option B .