The pH of `D_(2)O` and `H_(2)O` at 298 K is

Calculate the enthalpy change in the reaction, 4NH_(3) + 3O_(2) to 2N_(2) + 6H_(2) O at 298 K,given that the enthalpy of formation at 298 K for NH_(3) and H_(2)O are - 46.0 and - 286.0 KJ mol^(-1) respectively.

Of the two solvent H_(2)O and D_(2)O, NaCl dissolves

The standard reduction potential of the reaction H_(2)O+e^(-) rarr 1/2 H_(2) +OH , at 298 K, is

If S + O_(2) rarr SO_(2), Delta H = -298.2 kJ SO_(2) + (1)/(2) O_(2) rarr SO_(3), Delta H = -98.2 kJ SO_(3) + H_(2)O rarr H_(2)SO_(4), Delta H = -130.2 kJ H_(2) + (1)/(2) O_(2) rarr H_(2)O, Delta H = -287.3 kJ the enthalpy of formation of H_(2)SO_(4) at 298 K will be