Question

Suppose 200.0 mL of 1.00 M HCl and 200.0 mL of 1.00 M NaOH, both initially at 21.0°C, are mixed in a thermos flask. When the reaction is complete, the temperature is 27.8°C. Assuming that the solutions have the same heat capacity as pure water, compute the heat released (in kJ).

Answer 1

-11.4 kJ of heat are released

Step-by-step explanation:

Balanced reaction-
`HCl+NaOH\rightarrow NaCl+H_(2)O`

According to balanced equation, 1 mol of HCl is neutralized by 1 mol of NaOH.

So, 200 mL of 1 M HCl will be completely neutralized by 200 mL of 1 M NaOH.

Let's assume both HCl and NaOH solutions have densities equal to density of pure water.

We know, density = (mass)/(volume)

Density of pure water = 1 g/mL

So mass of both solutions =
`(200* 1)g` = 200 g

So, mass of mixture = (200+200) g = 400 g

Hence, Amount of heat released =
`(m_(mixture)* C_(mixture)* \Delta T_(mixture))`

Where, m represents mass, C represents heat capacity and
`\Delta T` represents change in temperature

Let's assume that heat capacity of mixture is also equal to heat capacity of water.

Heat capacity of water = 4.186 J/(g.
`^(0)\textrm{C}`)

So, amount of heat released =
`[400g* 4.186J.g^(-1).^(0)\textrm{C}^(-1)*(27.8-21.0)^(0)\textrm{C} ]` =
`1.14* 10^(4)`J=11.4 kJ

As acid-base neutralization reactions are exothermic therefore-

Heat released = -11.4 kJ