Question

A calorimeter is an insulated device in which a chemical reaction is contained. By measuring the temperature change, ΔT, we can calculate the heat released or absorbed during the reaction using the following equation: q=specific heat×mass×ΔT Or, if the calorimeter has a predetermined heat capacity, C, the equation becomes q=C×ΔT At constant pressure, the enthalpy change for the reaction, ΔH, is equal to the heat, qp; that is, ΔH=qp but it is usually expressed per mole of reactant and with a sign opposite to that of q for the surroundings. The total internal energy change, ΔE (sometimes referred to as ΔU), is the sum of heat, q, and work done, w: ΔE=q+w However, at constant volume (as with a bomb calorimeter) w=0 and so ΔE=qv.

Part A

A calorimeter contains 30.0 mL of water at 15.0 ∘C . When 1.50 g of X (a substance with a molar mass of 66.0 g/mol ) is added, it dissolves via the reaction X(s)+H2O(l)→X(aq) and the temperature of the solution increases to 30.0 ∘C . Calculate the enthalpy change, ΔH, for this reaction per mole of X. Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(g⋅∘C)], that density of water is 1.00 g/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings.

Answer 1

A. mass of solution = 30g+1.5g = 31.5g [water density 1gm/ml]

enthalpy change because of addition of 1.5g of X in 30ml of water is given as

(T₂=303k, T₁=288k)

  ΔH=ms ΔT = 31.5 x 4.184 x (303-288) = 31.5 x 4.184 x 15 = 1976.94J {specific heat of solution =specifif heat of water as per question}

no of moles of solute dissolved in the solution = 1.50/66 = 0.0227moleX.

so, 0.0227 mole X------> 1976.94J

1 mole X --------------> ?

applying cross multiplication we get, 1976.94x1/0.0227 =87089.86J/mole = 87.089KJ/mole