Calculate the total number of calories produced in the reaction.

Question

B. ΔH_solution for neutralization of HCl_(aq) and NaOH_(aq):

Volume of HCl: 0.0500 L

Volume of NaOH: 0.0500 L

Volume total: 0.1000 L

Molarity of HCl: 2.00 M

Molarity of NaOH: 2.00 M

ΔT for reaction B: 13.0 °C

Answer 1

HCl(aq) + NaOH(aq) ==> NaCl(aq) + H2O(l) ... balanced equation

q = mC∆T

q = heat = ?

m = mass = 0.1000 L x 1000 ml / L x 1.0 g / ml = 100 g (assuming a density of 1.0 g/ml)

C = specific heat = 1 cal/gº (assuming it is the same as that of water)

∆T = change in temperature = 13.0º

Solving for q, we have q = (100 g)(1 cal /gº)(13.0º)

q = 1300 cal

This is the heat generated for the given reaction. The given reaction contains 0.05 L x 2.00 mol/L of HCl and NaOH, so it contains 0.100 moles of each reactant. We want to express ∆H_rxn on a per mole basis.

∆H_rxn = 1300 cal / 0.100 mols = 13,000 cal/mol = 13.0 kcal/mol