Answer to Question #91541 in Physical Chemistry for Eric De Torres

Question

A student mixed 100.0 mL of 1.50 mol/L sulfuric acid with 200.0 mL of 1.50 mol/L sodium hydroxide. Both solutions were at 19.670C initially and the highest temperature reached by the reaction mixture was 34.060C. Calculate the molar enthalpy (deltaH) of neutralization for sulfuric acid.

Answer 1

Solution.

$$Q(solution) = (m(acid) + m(hydroxide)) \times c(H2O) \times (t2-t1)$$ $$m(acid) = C(acid) \times V(acid) \times M(acid)$$ $$m(hydroxide) = C(hydroxide) \times V(hydroxide) \times M(hydroxide)$$

Q(solution) = (14.7+12)*4.18*(34.06-19.67) = 1606.01 J

$$Q(reaction) = -Q(solution)$$

$$\Delta H = \frac{Q(reaction)}{n(lim)}$$

n(lim) is the limiting reagent in the reaction.

n(acid) = 0.15 mole

n(hydroxide) = 0.3 mole

Acid is limiting reagent.

$$\Delta H = \frac{-1606.01}{0.15}$$

$$\Delta H = -10706.73 J$$

Answer:

$$\Delta H = -10706.73 J$$