Question

PART A

A calorimeter contains 27.0 mL of water at 14.5 ∘C . When 1.70 g of X (a substance with a molar mass of 75.0 g/mol) is added, it dissolves via the reaction

X(s)+H2O(l)→X(aq)

and the temperature of the solution increases to 28.5 ∘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.

PART B

Consider the reaction

C12H22O11(s)+12O2(g)→12CO2(g)+11H2O(l)

in which 10.0 g of sucrose, C12H22O11, was burned in a bomb calorimeter with a heat capacity of 7.50 kJ/∘C. The temperature increase inside the calorimeter was found to be 22.0 ∘C. Calculate the change in internal energy, ΔE, for this reaction per mole of sucrose.

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

THANKS! (Detailed Steps Plz)

Answer 1

Part A

H = m C_p T for the substance X and water.

Now here m = mass of water + mass of substance X

= (Volume of water x density of water) + mass of substance X

= 27 X1 + 1.70 = 28.70 gm

C_p = 4.18 J/(gm ˚ C)

T = 28.5 - 14.5 = 14 ˚ C

H = 28.7 X 4.18 X 14 = 1679.52 J

Now mole of subtance X = mass / molar mass = 1.70 / 75 =0.02

H_solution = 1679.52 / 0.02 J/mole = 83976 J/mole = 83.98 KJ/mole

Part B

Here E = - m C_vT

m = 10 g

C_v = 7.5 KJ/gm ˚ C

T = 22 ˚ C

E = - 10 X 7.5 X 22 = -1650 KJ

Molar mass of C₁₂H₂₂O₁₁ = (12*12)+(22*1)+(11*16) = 144+22+176 = 342 g/mole

mole of Sucrose = mass / molar mass = 10 / 342 = 0.03

E_rxn = -1650 / 0.03 = - 55000 KJ/mole

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