A calorimeter contains 29.0 mL of water at 13.0 C . When 2.40 g of X (a substanc

Question

A calorimeter contains 29.0 mL of water at 13.0 C . When 2.40 g of X (a substance with a molar mass of 68.0 g/mol ) is added, it dissolves via the reaction

X(s)+H2O(l)X(aq)

and the temperature of the solution increases to 26.5 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.

Explanation / Answer

The data given in the question is as follows:
Mass of reactant = 2.40g
moles of reactant = m / molar mass = 2.40 / 68 = 0.035 moles
Mass of solution (m ) = 29 g [As density of water is 1.00 g/mL]
Intital temperature T1 = 13.0 °C
Final temperature T2 = 26.5 °C
Change in temperature T = T2- T1 = 26.5 - 13.0 = 13.5 °C
Specific heat capacity of water Cp = 4.18 J/g°C

The formula for generated heat can be written as follows:
Q = m x Cp x T ……….. (1)

Plug all the above data values in formula (1) and we get
Q = 29 x 4.18 x 13.5 j
Q= 1636.47 J = 1.636 kJ

Now the relation between Change in enthalpy and heat generated is as follows:
H = Q / number of moles
H = 1.636 kJ / 0.035 mol
H = 46.74 kJ/ mol
This reaction is producing energy which is raising the temperature and causing dissolution of X in water so it a exothermic reaction for which H is always negative
So our answer will be H = -46.74 kJ/ mol

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