A calorimeter contains 17.0 mL of water at 12.5 C . When 2.00 g of X (a substanc

Question

A calorimeter contains 17.0 mL of water at 12.5 C . When 2.00 g of X (a substance with a molar mass of 57.0 g/mol ) is added, it dissolves via the reaction

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

and the temperature of the solution increases to 30.0 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.

The change in H is=

Heat released by X raises the temperature of water and X from 12.5 oC to 30oC
Mass of water = Volume * density = 57 mL*1 g/mL = 57 g
mass of x =2.00 g
Heat required = (m1+m2)*C*(Tf-Ti)
  

I'm confused

Explanation / Answer

The heat released by substance dissolution in water causes the rise of temperature. We can write:

Q = m*c*T

Where:

Q = amount of heat transferred,
m = mass of water = 57.0 g (57.0 mL x 1.00 g/mL = 22.0 g)
C = specific heat of the solution [4.18 J/(g*C],
T = temperature change (Tfinal – Tinitial) [ 17.5]

As T >0, heat is absorbed by the solution and therefore Q has a positive value. This means that the dissolution of the substance X in water is an exothermic reaction:

H = - Q

Introducing the values in Q expression:

Q = 57*4.18*(30.0 – 12.5) = 4169.55 J

Calculate the number of moles of the substance (moles = mass in grams/molar mass):

Moles: 2 / 57 = 0.035 moles

H = -Q/moles = - 4169.55 /0.035 = -119130 J/mol = -119.13 kJ/mol

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