What is the enthalpy change(Q) when 60.0 g of NaOH is dissolved in one litre of water, given that the temperature of the solution increased by 15.8 °C?

ΔH = -44.029 KJ/mol

Explanation:

We are told that 60.0 g of NaOH is dissolved in one litre of water.

From conversion, 1 litre = 1000 g

From tables, specific heat capacity of water; c = 4.18 J/g/°C

We are given ΔT = 15.8 °C

Formula for the heat absorbed by the solution is given as;

Q = mcΔT

Where;

Q = Heat gained by the water

m = Mass of the water

c = Specific heat of water

ΔT = change in temperature

Thus;

Q = 1000 × 4.18 × 15.8

Q = 66044 J

The solution absorbed 66044 J and therefore it means that the dissolution of the salt gave off 66044 J.

Thus;

Enthalpy is; ΔH = -66044 J

Now, Mass of NaOH is 60 g.

We know that molar Mass of NaOH is 39.997 g/mol. Thus;

Converting to moles, we have;

n = 60/39.997

n = 1.5 mol

Now, enthalpy when 1.5 moles of NaOH are dissolved in water will be;

ΔH = -66044 J/1.5 moles

ΔH = -44029.33 J/mol

ΔH = -44.029 KJ/mol

c. 75.7

i think that is the right

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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