Question

Calcium carbonate reacts with hydrochloric acid to form calcium chloride, carbon dioxide, and water: CaCO₃ (s) + 2 HCl(aq) → CaCl₂ (aq) + CO₂(g) + H₂O(l)

When 20.0 g of calcium carbonate is mixed with excess hydrochloric acid to form 150.0 mL of solution in a coffee-cup calorimeter, the temperature changes from 24.55 °C to 29.05 °C. Find ΔH for the reaction per mole of calcium carbonate. Assume that the density of the solution is 1.00 g/mL and that its specific heat capacity is 4.18 J/g °C.

a. -1.42 x 10⁵ J/mol CaCO₃
b. 1.42 x 10⁵ J/mol CaCO₃
c. -2.82 x 10³ J/mol CaCO₃
d. 2.82 x 10³ J/mol CaCO₃

Answer 1

To find ΔH for the reaction per mole of calcium carbonate, you need to calculate the heat transferred during the reaction. The heat transferred can be calculated using the equation: q = m·c·ΔT. Therefore, the correct answer is option c. -2.82 x 10³ J/mol CaCO₃.

Step-by-step explanation:

To find ΔH for the reaction per mole of calcium carbonate, you need to calculate the heat transferred during the reaction.

The heat transferred can be calculated using the equation:

q = m·c·ΔT

Where:

• q is the heat transferred in joules
• m is the mass of the solution in grams
• c is the specific heat capacity of the solution in J/g °C
• ΔT is the change in temperature in °C

In this case, the mass of the solution is given as 150 g (density = 1.00 g/mL x 150 mL = 150 g). The specific heat capacity of the solution is given as 4.18 J/g °C. The change in temperature is 29.05 °C - 24.55 °C = 4.50 °C.

Substituting these values into the equation:

q = 150 g x 4.18 J/g °C x 4.50 °C = 2823 J

Since the reaction is exothermic, ΔH is equal to the negative of the heat transferred, so ΔH = -2823 J/mol.

Therefore, the correct answer is option c. -2.82 x 10³ J/mol CaCO₃.