If the standard enthalpy change ( _ r H ^ ) for reaction H ₂( ~g )+ Br ₂( l ) 2 HBr ( g ) is -72.8 ~kJ , the standard enthalpy of formation ( _ f H ^ ) of HBr ( g ) (in . kJ mol ⁻¹ ) is

Step-by-step solution

aligned & H _ f ^ ( H ₂ gas )=0.0 ~kJ ~mol ⁻¹, H _ f ^ ( Br ₂ liquid) . & =0.0 ~kJ ~mol ⁻¹ & H _ T ^ =-72.8 ~kJ for 2 moles of HBr . & H _ r ^ = [2 H _ f ^ ( H Br gas )- ( H _ f ^ ( H ₂ gas ) ) & .+ ( H _ f ^0 ( Br ₂ liq. ) ) ] & -72.8=2 H _ f ^ ( HBr gas )-(0.0+0.0) & =2 H _ f ^ ( HBr gas ) & H _ f ^ ( HBr gas )= -72.8 2 =,-36.4 ~kJ ~mol ⁻¹ . & aligned