Question

You react 0.017 mol of solid metal with HCl in a coffee cup calorimeter (reaction shown below). The calorimeter has 100 mL of water in it, and the temperature of the water increases by 3.81°C. The calorimeter has a heat capacity of 40.4 J/°C. What is the enthalpy of the reaction in terms of kJ per mol of the metal (your answer should be NEGATIVE, remember to convert from J to kJ, specific heat capacity of water is 4.184 J/g-°C)?

M(s) + 2 HCl (aq) MCl2 (aq) + H2 (g)

M = metal

Answer 1

To calculate the enthalpy of the reaction in terms of kJ per mol of the metal, we can use the following formula:

q = -mCΔT

where q is the heat absorbed by the water and the calorimeter, m is the mass of the water, C is the heat capacity of the calorimeter, and ΔT is the change in temperature of the water.

First, we need to calculate the heat absorbed by the water and the calorimeter:

q = (100 g) x (4.184 J/g-°C) x (3.81°C) + (40.4 J/°C) x (3.81°C)
q = 1657.4 J

Next, we need to calculate the moles of HCl used in the reaction. From the balanced chemical equation, we can see that 2 moles of HCl react with 1 mole of the metal, so:

moles of HCl = 2 x moles of metal = 2 x 0.017 mol = 0.034 mol

Finally, we can calculate the enthalpy of the reaction per mole of the metal:

ΔH = -q / moles of metal
ΔH = -(1657.4 J) / (0.017 mol)
ΔH = -97,494 J/mol

To convert to kJ/mol, we divide by 1000:

ΔH = -97.494 kJ/mol

Therefore, the enthalpy of the reaction is -97.494 kJ/mol of the metal. Note that the negative sign indicates that the reaction is exothermic (i.e. heat is released).