Question

A sample of K(s) of mass 2.720 g undergoes combustion in a constant volume calorimeter at 298.15 K. The calorimeter constant is 1849 J K−1, and the measured temperature rise in the inner water bath containing 1439 g of water is 1.60 K.

Part A

Calculate ΔU∘f for K2O.

Express your answer to three significant figures and include the appropriate units.

Part B

Calculate ΔH∘f for K2O.

Express your answer to three significant figures and include the appropriate units.

Answer 1

A) The ΔU° of
`K_2O` is -361 kJ/mol.

B) The ΔH° of
`K_2O` is -362 kJ/mol.

Step-by-step explanation:

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

`q=[q_1+q_2]`

`q=[c_1* \Delta T+m_2* c_2* \Delta T]`

where,

q = heat released by the reaction

= heat absorbed by the calorimeter

`q_2` = heat absorbed by the water

`c_1` = specific heat of calorimeter =
`1849 J/K`

`c_2` = specific heat of water =
`4.184J/gK`

`m_2` = mass of water = 1439 g

`\Delta T` = change in temperature =
`1.60 K`

Now put all the given values in the above formula, we get:

`q=[(1849 J/K * 1.60 K)+(1439 g * 4.184J/gK* 1.60 K)]`

`q= 12,591.64 J`

`4K(s)+O_2(g)\rightarrow 2K_2O(g)`

`2K(s)+(1)/(2)O_2(g)\rightarrow K_2O(g)`

m = mass of substance= 2.50 g

n = moles of substance

`n=(2.720 g)/(39 g/mol)=0.06974 mol`

According to reaction , 2 moles of K gives 1 mole of
`K_2O`

Then 0.06974 moles of K will give:

`(1)/(2)* 0.06974 mol=0.03487 mol` of
`K_2O`

12,591.64 Joules of energy was released when 0.03487 mol of
`K_2O` are formed.

So, the enthaply of formation of
`K_2O`: ΔU°

`=(-12,591.64 J)/(0.03487 mol)=-361,102.38 J/mol=-361.10 kJ/mol`

The ΔU° of
`K_2O` is -361 kJ/mol.

`\Delta H^o =\Delta U^o +\Delta n_gRT`

`\Delta n_g` = moles of gases on RHS - moles of gasses on LHS

`= -361,102.38 J/mol +(0-(1)/(2))* 8.314 J/mol* 298.15 K`

`\Delta H^o=-362 kJ/mol`

The ΔH° of
`K_2O` is -362 kJ/mol.