Given the reactions (1) and (2) below, determine (i) ΔrHimage and ΔrUimage for reaction (3), (ii) ΔfHimage for both HCl(g) and H2O(g), all at 298 K. (1) H2(g) + Cl2(g) → 2 HCl(g) ΔrHimage = −184.62 kJ mol−1 (2) 2 H2(g) + O2(g) → 2 H2O(g) ΔrHimage = −483.64 kJ mol−1 (3) 4 HCl(g) + O2(g) → 2 Cl2(g) + 2 H2O(g)

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation.

The given chemical reactions are:

(1) [tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g)[/tex]; [tex]\Delta H^o_1=-184.62kJ/mol[/tex]

(2) [tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(g)[/tex]; [tex]\Delta H^o_2=-483.64kJ/mol[/tex]

(3) [tex]4HCl(g)+O_2(g)\rightarrow 2Cl_2(g)+2H_2O(g)[/tex]; [tex]\Delta H^o_3=?[/tex]

Now we have to determine enthalpy change for reaction 3. We are reversing the equation (1) and multiplying by 2 and then adding both the equations, we get the chemical reaction 3.

(1) [tex]4HCl(g)\rightarrow 2H_2(g)+2Cl_2(g)[/tex]; [tex]\Delta H^o=2\times 184.62kJ/mol=369.24kJ/mol[/tex]

(2) [tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(g)[/tex]; [tex]\Delta H^o=-483.64kJ/mol[/tex]

The expression for enthalpy change for reaction 3 is:

[tex]\Delta H^o_3=\Delta H^o_1+\Delta H^o_2[/tex]

[tex]\Delta H^o_3=369.24+(-483.64)[/tex]

[tex]\Delta H^o_3=-114.4kJ/mol[/tex]

Thus, the change in enthalpy for reaction 3 is -114.4 kJ/mol.

Now we have to calculate the change in internal energy for reaction 3.

Formula used :

[tex]\Delta U=\Delta H-\Delta n_gRT[/tex]

[tex]\Delta H[/tex] = change in enthalpy

[tex]\Delta U[/tex] = change in internal energy

[tex]\Delta n_g[/tex] = change in moles

R = gas constant = 8.314 J/mol.K

T = temperature = 298 K

[tex]\Delta n_g[/tex] = change in moles

Change in moles = Number of moles of product side - Number of moles of reactant side

According to the reaction:

Change in moles = (2+2) - (4+1) = 4 - 5 = -1 mole

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

[tex]\Delta U=(-114.4kJ/mol)-(-1mol\times 8.314J/mol.K\times 298K)[/tex]

[tex]\Delta U=(-114.4\times 1000J/mol)-(-1mol\times 8.314J/mol.K\times 298K)[/tex]

[tex]\Delta U=-111.9kJ/mol[/tex]

Thus, the change in internal energy for reaction 3 is -111.9 kJ/mol.