Question

A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions were initially at 35.0°C and the temperature of the resulting solution was recorded as 37.0°C, determine the DH°rxn (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HCl. Assume 1) that no heat is lost to the calorimeter or the surroundings, and 2)that the density(1.00 g/mL) and the specific heat capacity (4.18 J/g oC) of the resulting solution are the same as water.

A.-16.7 kJ/mol NaOH

B. -27.9 kJ/mol NaOH

C. -169 kJ/mol NaOH

D. -34.4 kJ/mol NaOH

E. -55.7 kJ/mol NaOH

Answer 1

Since temperature of the solution rises, the reaction is exothermic in nature and therefore ∆H° is negative.

ULUI. * Total volume o solution = (100.0 + 100:o) mL = 200.0 ml Mass of solution = v*d = 200.0 ml x 1.00 g Now, heat change, Q = 200. g. = mass of solution & heat capacity of solution * AT = 200.9 x 4.18 Jo, x(37:0–35.0) = 1672 I 0.300 met 0.300 mol x 10 1000 m2 Tx 1000 ml Moles - = NaOH Now, 0.03 mol mol = 0.03 mol NaOH → NaOH → 1672 I of heat 1672 J 0.03 = 55.7 kJ So, atrxn = |- 55.7 kJ mol