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 get that answer.
You have 25.0 mL x 1.00 M = 25 mmols or 0.025 mols HBr and KOH reacting.
How much heat is generated? q = mcdT. m = mass = 25 + 25 = 50 grams. c = 1 cal/g. dT = delta T = 31.4-24.5 = 6.9 degrees.so q = 345 cal per 0.025 mols. How much is that for 1 mol? Remember this is an exothermic reaction so dH will be negative. Post your work/thoughts if you have more trouble.
thank you very much
To calculate the enthalpy of reaction for this chemical change, you need to use the equation q = mCΔT, where q is the heat evolved or absorbed by the reaction, m is the mass of the solution, C is the specific heat capacity of the solution, and ΔT is the change in temperature.
First, let's calculate the mass of the solution. Since the density of each solution is 1.00 g/mL, the mass of 25.0 mL of each solution is 25.0 g.
Next, we can calculate the heat evolved (q) using the equation q = mCΔT. In this case, we are assuming that the specific heats of each solution are 1.00 cal/g oC, so C = 1.00 cal/g oC.
The change in temperature (ΔT) is given by the final temperature (31.4 oC) minus the initial temperature (24.5 oC), which is ΔT = 31.4 oC - 24.5 oC = 6.9 oC.
Now we have all the values to calculate the heat evolved (q). Plugging in the values into the equation:
q = (25.0 g) * (1.00 cal/g oC) * (6.9 oC) = 1725 cal
Next, we need to convert the heat evolved from calories to kilocalories. Since there are 1000 calories in 1 kilocalorie, the heat evolved is:
q = 1725 cal / 1000 = 1.725 kcal
Finally, to calculate the enthalpy of reaction, we need to find the number of moles of the limiting reactant, which in this case is HBr. The volume of HBr used is 25.0 mL, which is equivalent to 0.025 L. The molarity (M) of HBr is given as 1.00 M. Therefore, the number of moles of HBr is:
moles HBr = Molarity * Volume = 1.00 mol/L * 0.025 L = 0.025 mol
Since the reaction produces 1 mol of HBr, the enthalpy of reaction for this chemical change is:
Enthalpy of reaction = q / moles HBr = 1.725 kcal / 0.025 mol = -13.8 kcal/mol
Therefore, the enthalpy of reaction for this chemical change is -13.8 kcal/mol.