The heat evolved in the reaction of hydrobromic acid with potassium hydroxide was determined using a coffee coup calorimeter. The equation for the reaction is:

HBr(aq) + KOH(aq) -------------> KBr(aq) + H2O(l)

When 25.0 mL of 1.00 M HBr at a temperature of 24.5oC was quickly mixed with 25.0 mL of 1.00 M KOH, also at 24.5 oC, the temperature rose to 31.4 oC. Calculate the enthalpy of reaction for this chemical change. Assume that the specific heats of each solution a re 1.00 cal/g oC and that the densities of each solution are 1.00 g/mL.

This is Joe again the answer should be -13.8kcal/mol but i don't know how to this get this answer.

The heat evolved in the reaction of hydrobromic acid with potassium hydroxide was determined using a coffee coup calorimeter. The equation for the reaction is:

HBr(aq) + KOH(aq) -------------> KBr(aq) + H2O(l)

When 25.0 mL of 1.00 M HBr at a temperature of 24.5oC was quickly mixed with 25.0 mL of 1.00 M KOH, also at 24.5 oC, the temperature rose to 31.4 oC. Calculate the enthalpy of reaction for this chemical change. Assume that the specific heats of each solution a re 1.00 cal/g oC and that the densities of each solution are 1.00 g/mL.

The answer should be -13.8kcal/mol but i don't know how to this get this answer.

I answered this above. The answer of -14.8 kcal/mol is correct.

See post above with Tom.

That's a typo. It should be -13.8 kcal/mol. See above under post by Tom for my response.

To calculate the enthalpy of reaction for this chemical change, we will use the equation:

q = m × C × ΔT

Where:
q is the heat evolved or absorbed by the reaction
m is the mass of the solution (in grams)
C is the specific heat of the solution (in cal/g°C)
ΔT is the change in temperature (in °C)

First, let's find the mass of the solution:
Density = mass/volume

Since the density of the solution is 1.00 g/mL, the mass of each solution is equal to its volume.

Mass of HBr solution = 25.0 mL = 25.0 g
Mass of KOH solution = 25.0 mL = 25.0 g

Next, let's calculate the heat evolved in the reaction using the equation mentioned above. We'll calculate q for both solutions and then sum them up:

For the HBr solution:
qHBr = mHBr × CHBr × ΔTHBr

Here:
mHBr = 25.0 g (mass of HBr solution)
CHBr = 1.00 cal/g°C (specific heat of HBr solution)
ΔTHBr = Tf - Ti (change in temperature of HBr solution)

Tf = final temperature = 31.4°C
Ti = initial temperature = 24.5°C

ΔTHBr = 31.4°C - 24.5°C = 6.9°C

Now, let's calculate qHBr:
qHBr = (25.0 g) × (1.00 cal/g°C) × (6.9°C)

Repeat the same steps for KOH solution:

For the KOH solution:
qKOH = mKOH × CKOH × ΔTKOH

Here:
mKOH = 25.0 g (mass of KOH solution)
CKOH = 1.00 cal/g°C (specific heat of KOH solution)
ΔTKOH = Tf - Ti (change in temperature of KOH solution)

ΔTKOH = 31.4°C - 24.5°C = 6.9°C

Now, let's calculate qKOH:
qKOH = (25.0 g) × (1.00 cal/g°C) × (6.9°C)

Finally, let's sum up the heat evolved in both solutions:
q_total = qHBr + qKOH

This q_total represents the heat evolved by the reaction. However, to calculate the enthalpy of reaction, we need to divide q_total by the number of moles of limiting reactant involved.

To determine the number of moles of the limiting reactant, we can use the stoichiometry of the balanced equation:

1 mole of HBr reacts with 1 mole of KOH.

In this case, we have taken equal volumes of both solutions (25.0 mL), so the number of moles of HBr and KOH will be equal.

Moles of HBr = Moles of KOH = (1.00 M) × (0.0250 L) = 0.025 moles

Now, divide q_total by the number of moles of the limiting reactant:
ΔH = q_total / (moles of limiting reactant)

ΔH = q_total / 0.025 moles

Finally, substitute the calculated values into the equation and solve for ΔH to find the enthalpy of reaction.