Answer
i) $\Delta_{r}$$H^o$=-114.4KJ/mol
$\Delta_{r}$$U^o$=-111.9KJ/mol
ii) $\Delta_{f}$$H^o$(HCl)=-92.31KJ/mol
$\Delta_{f}$$H^o$($H_{2}$O)=-241.82KJ/mol
Work Step by Step
i)
Step 1: Let's determine how to obtain reaction equation (3) by processing reaction equations (1) & (2) with arithmetic methods. Each time, we process $\Delta_{r}$$H^o$ of a reaction in the same way arithmetically as we process the reaction equation:
-- Notice in reaction equation (3) HCl is on the left and in reaction equation (1) it is on the right. So first let's flip the reactant and product in reaction equation (1) and $\Delta_{r}$$H^o$ of the reaction equation formed would be the opposite number of $\Delta_{r}$$H^o$ of reaction equation (1) :
2HCl(g) -->$H_{2}$(g)+$Cl_{2}$(g) (equation 4) $\Delta_{r}$$H^o$=-(-184.62 KJ/mol)=184.62 KJ/mol
-- Then, notice the coefficient for HCl in reaction equation (3) is 4, so we multiply 2 on reaction equation (4) and multiply 2 on its $\Delta_{r}$$H^o$ as well:
4HCl(g) -->2$H_{2}$(g)+2$Cl_{2}$(g) (Equation 5) $\Delta_{r}$$H^o$=184.62$\times$2=369.24KJ/mol
-- Finally, we directly add up reaction equations (2) and (5) as 1 molecule $O_{2}$ is on left of reaction equation (2) as in reaction equation (3), and add up $\Delta_{r}$$H^o$ for reaction equations (2) and (5) as well:
4HCl(g)+ $O_{2}$(g)+2$H_{2}$(g)-->2$H_{2}$(g)+2$Cl_{2}$(g)+2$H_{2}$O(g)
(Cancel out 2$H_{2}$) 4HCl(g)+ $O_{2}$(g)-->2$Cl_{2}$(g)+2$H_{2}$O(g) (Reaction Equation 3)
$\Delta_{r}$$H^o$=(-483.64)+369.24=-114.4KJ/mol
****$\Delta_{r}$$H^o$ of reaction equation (3) is -114.4KJ/mol.
Step 2: $\Delta_{r}$$U^o$=$\Delta_{r}$$H^o$ -$\Delta$(PV)=$\Delta_{r}$$H^o$ -V$\Delta$P (As V is constant for $\Delta$U)
Assume all gases to be ideal gas, and at constant V of the reaction system:
V$\Delta$P=($\Delta$n)RT
So $\Delta_{r}$$U^o$=$\Delta_{r}$$H^o$-($\Delta$n)RT
= -114.4KJ/mol-(2+2-4-1)$\times$(8.31$\times$$10^{-3}$KJ/(K·mol))$\times$298K
=-111.9 KJ/mol
ii) The enthalpy of formation $\Delta_{f}$$H^o$ for HCl (g) and $H_{2}$O(g) is the enthalpy for forming 1 mol of HCl (g) and 1 mol $H_{2}$O(g), and we notice the coefficients of HCl (g) and
$H_{2}$O(g) in reaction equations (1) & (2) are both 2. So $\Delta_{f}$$H^o$ for HCl (g) and $H_{2}$O(g) would be one half of $\Delta_{r}$$H^o$ for equations (1) & (2) respectively.
