Answer:
a. Lithium is in its standard state
b. [tex] Li (s) + \frac{1}{2} O_2 (g) + \frac{1}{2} H_2 (g)\rightarrow LiOH (s)[/tex]
c. [tex]-440 kJ/mol[/tex]
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:
- [tex]Li (s)[/tex]: solid lithium metal in its standard state;
- [tex]O_2 (g)[/tex]: oxygen gas as diatomic in its standard state;
- [tex]H_2 (g)[/tex]: hydrogen gas as diatomic in its standard state.
We obtain the following equation:
[tex] Li (s) + \frac{1}{2} O_2 (g) + \frac{1}{2} H_2 (g)\rightarrow LiOH (s)[/tex]
c. Firstly, write the equation for the enthalpy of formation of water using the guidelines in (b):
[tex]H_2 (g) + \frac{1}{2} O_2 (g)\rightarrow H_2O (l); \Delta H_1 = -286 kJ/mol[/tex]
Now given the equation:
[tex]LiOH (s) \rightarrow Li^+ (aq) + OH^- (aq); \Delta H_2 = -21 kJ/mol[/tex]
As well as:
[tex] Li (s) + \frac{1}{2} O_2 (g) + \frac{1}{2} H_2 (g)\rightarrow LiOH (s); \Delta H_3 = -485 kJ/mol[/tex]
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:
[tex]2 Li (s) + 2 H_2O (l)\rightarrow 2 Li^+ (aq) + 2 OH^- (aq) + H_2 (g); \Delta H_4[/tex]
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:
[tex]\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[/tex]
