Answer to Question #150191 in General Chemistry for Kirt Jones Laurente

Question

An experiment was performed to determine the enthalpy change, ΔH, for the reaction of 100.0 mL of 0.50 M HCl and 100.0 mL of 0.50 M NaOH solution. If both solutions were initially at 20.10°C and the temperature of the resulting solution was recorded as 23.40°C, how much heat(in joules) was evolved in the water of the solution? Specific heat capacity of solutions (cg) = 4.184 J/°C·­ g­. The density of water is 1.000 g/mL

Answer 1

Q150191

An experiment was performed to determine the enthalpy change, ΔH, for the reaction of 100.0 mL of 0.50 M HCl and 100.0 mL of 0.50 M NaOH solution. If both solutions were initially at 20.10°C and the temperature of the resulting solution was recorded as 23.40°C, how much heat(in joules) was evolved in the water of the solution? Specific heat capacity of solutions (Cs) = 4.184 J/°C*g. The density of water is 1.000 g/mL

Solution:

1) We have to find the amount of heat energy given by the reaction of HCl and NaOH. The reaction of HCl and NaOH is given as,

NaOH (aq) + HCl (aq) -----> NaCl (aq) + H₂O(l) + Q ;

where Q is the heat given out by the reaction. This heat energy Q will cause an increase in the temperature of the water. That is why we observe an increase in temperature from 20.10°C to 23.40°C.

2) 100.0 mL of 0.50 M HCl and 100.0 mL of 0.50 M NaOH solution is mixed.

Final volume after mixing both the solution = 100.0mL + 100.0 mL = 200.0 mL.

Consider the density of the final solution the same as that of water which is 1.000 g/mL. 

$density = \frac{mass }{volume};$ So, mass = density * volume .

mass, m = 1.000 g/mL * 200.0 mL = 200.0 grams.  

So now we have,

mass of solution after mixing, m = 200.0 grams

specific heat capacity of the solution, C_S = 4.184 J/°C·g

Initial temperature of the solution, T_I = 20.10 °C

Final temperature of the solution, T_F ( after mixing) = 23.40°C.

Change in Temperature, ∆T = T_F - T_I = 23.40°C - 20.10 °C = 3.3 °C.  

Formula :

The relation between heat energy and specific heat capacity is given as,

Q = m * C_S * ∆T;

Solution :

Substitute all the given information in the formula, we have

Q = 200.0 g *4.184 J/g^{-1 0}C^-1 * 3.3⁰C ;

Q = 2761.44 Joules. 

In question we are given the quantity with the least number of the significant figure is the molarity

0.50 M HCl and 0.50 M NaOH. They are in 2 significant figure, so our final answer must also be in

2 significant figures.

Hence 2800 Joules of heat energy will be given by the reaction of NaOH and HCl to the water of the solution.