Question

Can anyone help me with this please! my exam is day after tomorrow

Answer 1

a. Lithium is in its standard state

b.
`Li (s) + (1)/(2) O_2 (g) + (1)/(2) H_2 (g)\rightarrow LiOH (s)`

c.
`-440 kJ/mol`

Step-by-step explanation:

a. Elements in their standard states at room temperature and 1 atm pressure would have an enthalpy of formation of 0 kJ/mol. Lithium is metal at standard conditions, so its enthalpy of formation is 0 kJ/mol.

b. The equation representing the formation of a compound should following the rules below:

• strictly 1 mol of the product formed;
• all reactants in their standard states at room temperature and 1 atm pressure.

We should, hence, form 1 mol of LiOH from the following species:

• 
  `Li (s)`: solid lithium metal in its standard state;
• 
  `O_2 (g)`: oxygen gas as diatomic in its standard state;
• 
  `H_2 (g)`: hydrogen gas as diatomic in its standard state.

We obtain the following equation:

`Li (s) + (1)/(2) O_2 (g) + (1)/(2) H_2 (g)\rightarrow LiOH (s)`

c. Firstly, write the equation for the enthalpy of formation of water using the guidelines in (b):

`H_2 (g) + (1)/(2) O_2 (g)\rightarrow H_2O (l); \Delta H_1 = -286 kJ/mol`

Now given the equation:

`LiOH (s) \rightarrow Li^+ (aq) + OH^- (aq); \Delta H_2 = -21 kJ/mol`

As well as:

`Li (s) + (1)/(2) O_2 (g) + (1)/(2) H_2 (g)\rightarrow LiOH (s); \Delta H_3 = -485 kJ/mol`

Notice that multiplying reaction (2) by 2, multiplying reaction (3) by 2, multiplying reaction (1) by -2 (that is, multiplying by 2 and reversing it) and adding them together will yield the target equation:

`2 Li (s) + 2 H_2O (l)\rightarrow 2 Li^+ (aq) + 2 OH^- (aq) + H_2 (g); \Delta H_4`

According to Hess's Law, we will perform the same steps with the enthalpy values and we will add them to get the final enthalpy value:

`\Delta H_4 = 2\Delta H_2 + 2\Delta H_3 + (-2)\Delta H_1 = 2\cdot (-21 kJ/mol) + 2\cdot (-485 kJ/mol) - 2\cdot (-286 kJ/mol) = -440 kJ/mol`