Question

A student mixed 0.0100L of 0.500mol/L of KOH (aq) and 0.0100L of 0.500mol/L of H2SO4. The temperature increased by 4.00 degrees. Given Q=1.386kJ,

find molar entharpy of water formed in the reaction, in kJ/mol

How do I find the molar entharpy for the product of the reaction?

I want fast answers ASAP.. please.. help me..

Answer 1

Solution:

The reaction between H2SO4 and KOH to form water is expressed as,

H2SO4 + 2KOH = 2H2O + K2SO4

Thus, number of moles of H2O is equal to number of moles of KOH.

The number of moles of H2SO4 = Molarity x Volume

= 0.500 mol /L. x 0.0100 L = 0.005 mol

The number of moles of KOH = Molarity x Volume

= 0.500 mol/L x 0.0100 L = 0.005 mol

Hence, the number of moles of H2O = 0.005 mol

Thus,

q = m c ΔT

q = 20 g x 4.184 J /°C g x 4 °C

= 334.72 J

(Since, mass of KOH and H2SO4 is equal to its volume because density of solution is considered as 1g/mL. Mass = 0.010 L + 0.010 L = 20 mL = 20 g )

Thus,

ΔH = - q /n = - 334.72 J /number of moles of water

ΔH = - 334.72 J /0.005 mol = - 66944 J/mol

ΔH = - 66.94 kJ/mol

Hence, molar heat of enthalpy (ΔH)for H2O = -66.94 kJ/mol

Total number of moles of product = moles of H2O and K2SO4

= 0.005 mol + 0.005 mol = 0.010 mol

Thus,

Molar enthalpy of products = - Q/number of moles

= - 1.386 kJ / 0.01 mol

= - 138.6 kJ/mol