Calculate the standard enthalpy change for the following reaction at 25 °C. HCl (g) + NaOH (s) - > NaCl (s) + H2O (l)

Standard enthalpy change could be calculated as summation of standard enthalpy of formation of products minus summation of standard enthalpy of formation of products.

The standard enthalpy of formation:

NaCl (s) = - 411.13 kJ/mol

H2O (l) = - 241.818 kJ/mol

HCl (g) = - 92.30 kJ/mol

NaOH (s) = - 425.93 kJ/mol

Since the coefficients of the equation is 1, proceed to calculation

Standard enthalpy change = [ (-411.13) + (-241.818) ] - [ (-425.93) + (-92.30) ] = - 134.718 kJ/mol