In order to verify the heat of reaction, a chemist mixes 1 liter of 1 M NaOH and

Question

In order to verify the heat of reaction, a chemist mixes 1 liter of 1 M NaOH and 1 liter of 1 M HCl. The initial temperature of the two solutions is 25°C. The specific heat of water is 4.184 J/g°C.

The enthalpy of reaction of the following is –58 kJ/mol.

HCl(aq) + NaOH(aq) ® NaCl(aq) + H2O(l)

A) Is this reaction exothermic or endothermic? (Will the temperature increase or decrease?)

B) Considering only the specific heat of the 2 liters of water and no heat transfer to/from surroundings, what is the final temperature of the solution (notice that the specific heat is in terms of mass, not volume)?

Explanation / Answer

It is an exothermic reaction, heat was lost to the water and it got warmer.

(B)

These are solutions, not pure water. The specific heat of water is 4.184 J/goC. Assume that these solutions are close enough to being like water that their specific heats are also 4.184 J/goC.

The density of water is 1.00 g/mL and even though these are solutions we can assume that they are close enough to water to have the same density.

Solution

q = mct

We have L's and we need grams.

Use density. (1 L + 1 L ) = 2 L of solution.

2000 mL X 1 g/mL = 2000 grams of solution. (m = V x D)

Also at constant pressure the heat flow (q) for the process is equal to the change in enthalpy defined by the equation:

H = q = mct

t = H / (mc)

      = (– 58000 J) / ((2000 g) x (4.184 J/g 0C))

      = - 6.93 0C

So, the final temperature = 25 0C - (-6.93 0C) = 31.93 0C.

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