Naoh(s) dissolves readily in water to form na+(aq) and oh-(aq). if 35.1 g of naoh is dissolved in 199 g of water in a coffee cup calorimeter, the temperature of the resulting solution increases from 23.0 oc to 62.6 oc. calculate the enthalpy change per mole of naoh dissolved. assume the specific heat capacity of the solution is 4.18 j/oc*g.

ΔH = -44,13 kJ/mol

Explanation:

For the following reaction:

NaOH(s) → Na⁺(aq) + OH⁻(aq)

It is possible to obtain the heat produced using:

Q = 4,18J/g°C×mass×ΔT

Where the mass is 35,1g + 199g = 234,1g

And ΔT is FinalT - InitialT → 62,6°C - 23,0°C = 39,6°C

Replacing:

Q = 38750 kJ = 38,75kJ

The enthalpy change is obtained using:

ΔH = -Q/moles of NaOH

Moles of NaOH are:

35,1g×= 0,878 moles

Thus:

ΔH = -38,75 kJ/0,878mol = -44,13 kJ/mol

I hope it helps!

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explanation:

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