When 25.0 mL of 0.25mol/L LiOH and 25.0 mL of 0.25 mol/L HCl are mixed together, the temperature warms 15.8 oC. What is the molar enthalpy of neutralization for LiOH?
"LiOH + HCl = LiCl + H2O"
n(LiOH) = n(HCl)
"m(solution) = V(solution) \\times \\rho"
m(solution) = 50 g
"Q = c \\times m \\times \\Delta T"
"\\Delta H =- Q"
"Q = 4200 \\times 0.05 \\times 15.8 = 3318 \\ J"
"\\Delta H = -3318 \\ J"
"\\Delta H(molar) = \\frac{\\Delta H}{n(LiOH)}"
"\\Delta H(molar) = \\frac{-3318}{0.25*0.025} = 530.9 \\ kJ\/mol"
Answer:
"\\Delta H(molar) = 530.9 \\ kJ\/mol"
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