previous | 5 of 12next Part A A calorimeter contains 28.0 mL ot water at 1 30 \"

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previous | 5 of 12next Part A A calorimeter contains 28.0 mL ot water at 1 30 "C. when 250 gof X (a substance wth a molar mass of 610 g/mal ) s assed, a dissolves via the reaction and the lemperature of the solution increases to 27.0° Calculate the enthalpy change, 1·for ths reacton per mole of X Assume that the specific heat of the resulting soludtion is equal to that of water [4.18 J/(g x)HO0X(a) "C)l, that density of walter is 1.00 g/mL and that no heat is lost to the calonmeter itsell, nor to the surroundings Express the change in enthalpy in kilojoules per mole to three significant tigures AH 40.1 kJ/mol Submit ints MyAnswers Give Up Review Part Incorrect; Try Again; 3 attempts remaining t9 144 6 8 9

Explanation / Answer

Part-A: The dissolution reaction of X(s) is

X(s) + H2O --- > X(aq), DeltaH = ?

Since the temperature of water increases, this reaction is an exothermic reaction.

Hence heat absorbed by water = Heat released during dissolution.

Mass of water, m = Vxd = 28 mL x (1g/mL) = 28 g

s for water = 4.18 J/g.DegC

Change in temperature, dT = 27.0 - 13.0 = 14 DegC

Heat absorbed by water, Q = mxsxdT = 28g x(4.18 J/g.DegC) x 14 DegC = 1638.6 J

Hence heat released by the dissolution of 2.5 g of X(s) = - 1638.6 J

Now heat released by the dissolution of 61.0 g ( = 1 mole) of X(s) = ( - 1638.6 J) x (61.0 g/ 2.5 g)

= - 39982 J/mol = - 40.0 KJ/ mol (answer)

Hence enthalpy change for the reaction per mole of X = - 40.0 KJ/ mol.

Part:B: C12H22O11(s) + 12O2(g) ---> 12 CO2(g) + 11H2O(l)

Molecular mass of sucrose = 342.29 g/mol

When 10.0 g of cucrose is heated, the increase in the internal energy of the bomb calorimeter

= CvxdT = (7.50 KJ/DegC) x 22.0 DegC = 165 KJ

Hence the change in internal energy when 1 mole ( = 342.29 g/mol) of sucrose is burnt

= 165 KJ x (342.29 g/mol / 10.0 g) = 5650 KJ/mol (answer)

Hence DeltaE = 5650 KJ/mol

Q.2: Part-A: Enthalpy of formation is defined as the enthalpy change when 1 mole of a substance is formed from the constituent elements inn their standard state.

For the following reactions DeltaH(rxn) is equals to DeltaHf

C(s) + O2(g) -- > CO2(g)

Na(s) + 1/2Cl2(g) -- > NaCl(s)

Part-B: C5H12(g) + 8O2(g) --- > 5CO2(g) + 6H2O(g)

Enthalpy of combustion for the above reaction can be calculated as

DeltaH(comb) = 5 x[DeltaHf(CO2,g)] + 6 x[DeltaHf(H2O,g)] - [DeltaHf(C5H12,g)] - 8 x[DeltaHf(O2,g)]

= 5 molx(-393.5 KJ/mol) + 6 mol x (-241.8 KJ/mol) - (-110.0 KJ/mol) - 0

Note: The numerical values are not clearly visible in the figure, put the correct one if entered wrong.

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