Question

Calculate the standard enthalpy change (ΔH°) for the reaction where 1 mol of solid sodium hydroxide (NaOH) dissolves in excess water to produce 1 mol of aqueous solution and releases heat. Given that the standard enthalpy of formation (ΔH°f) for NaOH(s) is -425.6 kJ/mol, the heat capacity of the aqueous solution is 4.18 J/gK, and the change in temperature (ΔT) is observed to be 8.0°C for 250 g of the solution.

Answer 1

To calculate the standard enthalpy change  H  for the reaction, we first need to determine the heat  q  released when 1 mol of NaOH dissolves in water. We can use the formula:q = mcTwhere m is the mass of the solution  250 g , c is the heat capacity of the solution  4.18 J/gK , and T is the change in temperature  8.0C .q =  250 g  4.18 J/gK  8.0 K q = 8,350 JSince the reaction releases heat, the value of q is negative:q = -8,350 JNow, we need to convert the heat released  q  to kJ/mol:q = -8,350 J   1 kJ / 1000 J  = -8.35 kJSince 1 mol of NaOH is dissolved, the enthalpy change for the reaction is equal to the heat released per mole:H = -8.35 kJ/molTherefore, the standard enthalpy change  H  for the reaction where 1 mol of solid sodium hydroxide  NaOH  dissolves in excess water to produce 1 mol of aqueous solution and releases heat is -8.35 kJ/mol.