Question

A total of 2.00 mol of a compound is allowed to react with water in a foam coffee cup and the reaction produces 126 gg of solution. The reaction caused the temperature of the solution to rise from 21.00 to 24.70 ∘C∘C. What is the enthalpy of this reaction? Assume that no heat is lost to the surroundings or to the coffee cup itself and that the specific heat of the solution is the same as that of pure water.

Answer 1

Answer : The enthalpy of this reaction is, 0.975 kJ/mol

Explanation :

First we have to calculate the heat produced.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = ?

m = mass of solution = 126 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $21.00^oC$

$T_2$ = final temperature = $24.70^oC$

Now put all the given values in the above formula, we get:

$q=126g\times 4.18J/g^oC\times (24.70-21.00)^oC$

$q=1948.716J=1.95kJ$

Now we have to calculate the enthalpy of this reaction.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 1.95 kJ

n = moles of compound = 2.00 mol

Now put all the given values in the above formula, we get:

$\Delta H=\frac{1.95kJ}{2.00mole}$

$\Delta H=0.975kJ/mol$

Thus, the enthalpy of this reaction is, 0.975 kJ/mol