Calculate the Delta H at 298 K for the reaction (1)/(2)N_(2)(g) + (3)/(2)H_(2)(g) rarr NH_(3)(g) given that Delta H for the formation of NH_(3) has a valve of -46.0 kJ mol^(-1) (R = 8.314 JK^(-1) mol^(-1)).

Calculate the Delta H at 298 K for the reaction (1)/(2)N_(2)(g) + (3)/

1.	Calculate the Delta H at 298 K for the reaction (1)/(2)N_(2)(g) + (3)/(2)H_(2)(g) rarr NH_(3)(g) given that Delta H for the formation of NH_(3) has a valve of -46.0 kJ mol^(-1) (R = 8.314 JK^(-1) mol^(-1)).
Answer» Solution :Using the eqn. `Delta H = Delta U + Delta nRT` `Delta H = -46.0 kJ MOL^(-1) = -46000 J mol^(-1), Delta N = 1 - 2 = -1, R = 8.31 JK^(-1) mol^(-1)`, `T = 298 K ""Delta H = ?` `Delta U = Delta H - Delta nRT` `= -46000 J mol^(-1) -[(-1) xx 8.314 JK^(-1) mol^(-1) xx 298 K]` `= 43523 J mol^(-1) = -43.52 kJ mol^(-1)` Change in the internal energy for this reaction `= -43.52 kJ mol^(-1)`.