Question

When 4.93 g of ammonium bromide (NH4Br) is dissolved in 103 g of water in a styrofoam calorimeter of negligible heat capacity, the temperature drops from 25.00 to 22.97 °C. Based on this observation, calculate q for the water and ΔH° for the process, assuming that the heat absorbed by the salt is negligible.

NH4Br(s) NH4+(aq) + Br- (aq)

The specific heat of water is 4.184 J °C-1 g-1.

Answer #1

The amount of heat lost by water is , q = mcdt

Where

m = mass of water = 103 g

c = specific heat capacity of water = 4.184 J °C-1 g-1

dt = change in temperature = initial - final

= 25.00 - 22.97 oC

= 2.03 oC

Plug the values we get q = 103x4.184x2.03

= 874.8 J

This much amount of heat is absorbed by 4.93 g of NH4Br

The amount of heat absorbed by 1 mol = 97.94 g of NH4Br is = ( 97.94x874.8)/4.93

= 17379.5 J

So ΔH° = 17379.5 J

= 17.379 x10^{3} J

= 17.379 kJ

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