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

water at room temperature is flowing through a pipe with pressure taps located 10 ft apart along one section of the pipe, which are connected to a mercury (s=13.6) manometer.