The enthalpy change ∆ H for the reaction N 2 ( g ) + 3 H 2 ( g ) → 2 NH 3 ( g ) is - 92 . 38 kJ at 298 K . What is ∆ U at 298 K ?

Given, N 2 ( g ) + 3 H 2 ( g ) → 2 NH 3 ( g ) . ∆ H = - 92 .38 kJ . T = 298 K . ∆ U = ? Formula used in the calculation is as follows, ∆ H = ∆ U + ∆ n g RT . ∆ U = ∆ H - ∆ n g RT . Where, ∆ H = Enthalpy change ∆ U = Internal energy change ∆ n g = Change in number of gas species (Number of moles). ∆ n g = 2 - 4 = - 2 . R = Gas Constant 8 . 314 J K - 1 mol - 1 . T = Temperature. So, ∆ U = ∆ H - ∆ n g RT . ∆ U = - 92 .38 - - 2 8 .314 × 0 .001 × 298 . ∆ U = - 92 . 38 + 4 . 955 = - 87 . 425 kJ . [ ∵ Multiplied by 0 . 001 to Change joule into kJ ].

The internal energy change ∆ U for the reaction CH 4 ( g ) + 2 O 2 g → CO 2 ( g ) + 2 H 2 O ( l ) is – 885 kJ mol - 1 at 298 K . What is ∆ H at 298 K ?

Given, CH 4 ( g ) + 2 O 2 ( g ) → CO 2 ( g ) + 2 H 2 O l . ∆ U = - 885 kJ mol - 1 . T = 298 K . ∆ H = ? Formula used for the calculation is ∆ H = ∆ U + ∆ n g RT . Where, ∆ H = Enthalpy ∆ U = Internal Energy. ∆ n g = Change in number of gas species (Number of moles). ∆ n g = 1 - 3 = - 2 . R = Gas Constant ( 8 . 314 J K - 1 mol - 1 ). T = Temperature Change in enthalpy: ∆ H = ∆ U + ∆ n g RT . ∆ H = ( - 885 ) + ( - 2 ) 8 .314 × 0 .001 × 298 ∵ 1 J = 0 . 001 kJ . ∆ H = - 885 - 4 . 955 = - 889 . 955 kJ .