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

Question #91541
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.
1
Expert's answer
2019-07-10T02:52:55-0400

Solution.

Q(solution)=(m(acid)+m(hydroxide))×c(H2O)×(t2t1)Q(solution) = (m(acid) + m(hydroxide)) \times c(H2O) \times (t2-t1)m(acid)=C(acid)×V(acid)×M(acid)m(acid) = C(acid) \times V(acid) \times M(acid)m(hydroxide)=C(hydroxide)×V(hydroxide)×M(hydroxide)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)Q(reaction) = -Q(solution)

ΔH=Q(reaction)n(lim)\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.


ΔH=1606.010.15\Delta H = \frac{-1606.01}{0.15}

ΔH=10706.73J\Delta H = -10706.73 J

Answer:

ΔH=10706.73J\Delta H = -10706.73 J


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