Question

Part A: A calorimeter contains 32.0 mL of water at 12.5 ∘C . When 1.80 g of X (a substance with a molar mass of 72.0 g/mol ) is added, it dissolves via the reaction

X(s)+H2O(l)→X(aq)

and the temperature of the solution increases to 27.0 ∘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.

Express the change in enthalpy in kilojoules per mole to three significant figures.

Answer 1

Temperature change of water 12.5⁰C to 27⁰C

T = 27 - 12.5 = 14.8⁰C

specific of water = 4.18 J/ g ⁰C

mass of water = 32 ml = 32 gm

q = mass of water X  specific heat of H₂O_(l)  X  T

= 32 X 4.18 X 14.8

q = 1979.648J

H = 1979.648J for 1.80 gm of X substance

molar mass of X = 72 gm/mole then 1.80 gm of X = 1.80 / 72 = 0.025 mole

for 0.025 mole H = 1979.648J then for 1 mole H = 1979.648 X 1 / 0.025 = 79185.92 J

1000 J = 1 KJ then 79185.92 J = 79.186 KJ

tempreture water in this reaction increased after addition of X that mean heat is released thus H have negative sign

H = -79.186 KJ