Part A A calorimeter contains 35.0 mL of water at 14.0 C . When 2.40 g of X (a s

Question

Part A

A calorimeter contains 35.0 mL of water at 14.0 C . When 2.40 g of X (a substance with a molar mass of 67.0 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, H , 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/(gC) ], 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, E , for this reaction per mole of sucrose.

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

Explanation / Answer

m = 35 g of water, Tw = 14 °C

m = 2.40 g of X MWx = 67

mol = mass/MW = 2.4/67 = 0.03582 mol of X

Tf = 28.5°C

Calculate Enthalpy change

Q = m*C*(Tf-Ti) = (35+2.4)*4.184*(28.5-14)

Q = 2268.9832J

HRxn = -Q/mol = -2268.9832/(0.03582 )

HRxn = -63344.031 J/mol

HRxn = -63.344 kJ/mol

B.

mol = mass/MW = 10/342.2965 = 0.0292 mol of scurose

C = 7.5 kJ/°C

Qcalorimeter = C*dT = (7.5*22) = 165 kJ

HRxn = -Q/mol

HRxn = -165 /0.0292

HRxn = -5650.684 kJ/mol

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