Find `Delta_(r)H^@` for the reaction `4HCl(g) +O_(2)(g) rarr2Cl_(2)(g)+2H_(2)O(g)` at 300 K , Assume all gases are ideal.
Given: `H_(2)(g)+Cl_(2)(g)rarr2HCl(g) " " Delta_rH_(300)^(@)=-184.5 kJ`/mol
`2H_(2)(g)+O_(2)(g)rarr2H_(2)O(g)" " Delta_(r)H_(300)^(@)=-483kJ` /mol

Options

A. `111.5 ` kJ/mole
B. `-109.01 ` kJ/mole
C. `-111.5` kJ/mole
D. `-114` kJ/mole

Answer

To find the standard reaction enthalpy (Δ_rH^@) for the reaction:

\[ 4 \text{HCl}(g) + \text{O}_2(g) \rightarrow 2 \text{Cl}_2(g) + 2 \text{H}_2\text{O}(g) \]

at 300 K, we will use Hess's law and the provided enthalpy changes for the related reactions.

### Step 1: Write the given reactions and their Δ_rH values.

1. Formation of HCl:
\[ \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2 \text{HCl}(g) \quad \Delta_rH_{300}^@ = -184.5 \text{ kJ/mol} \]

2. Formation of water:
\[ 2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2\text{O}(g) \quad \Delta_rH_{300}^@ = -483 \text{ kJ/mol} \]

### Step 2: Reverse the first reaction to get HCl on the reactant side.

Reversing the first reaction gives:
\[ 2 \text{HCl}(g) \rightarrow \text{H}_2(g) + \text{Cl}_2(g) \quad \Delta_rH_{300}^@ = +184.5 \text{ kJ/mol} \]

### Step 3: Multiply the reversed reaction by 2.

To obtain 4 moles of HCl, we multiply the entire reversed reaction by 2:
\[ 4 \text{HCl}(g) \rightarrow 2 \text{H}_2(g) + 2 \text{Cl}_2(g) \quad \Delta_rH_{300}^@ = 2 \times 184.5 \text{ kJ/mol} = +369 \text{ kJ/mol} \]

### Step 4: Write the second reaction as is.

The second reaction remains unchanged:
\[ 2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2\text{O}(g) \quad \Delta_rH_{300}^@ = -483 \text{ kJ/mol} \]

### Step 5: Add the modified reactions.

Now, we add the modified reactions:
1. \( 4 \text{HCl}(g) \rightarrow 2 \text{H}_2(g) + 2 \text{Cl}_2(g) \) (Δ_rH = +369 kJ)
2. \( 2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2\text{O}(g) \) (Δ_rH = -483 kJ)

When we add these reactions, we get:
\[ 4 \text{HCl}(g) + \text{O}_2(g) \rightarrow 2 \text{Cl}_2(g) + 2 \text{H}_2\text{O}(g) \]

### Step 6: Calculate the total Δ_rH for the reaction.

Now, we can calculate the total Δ_rH for the desired reaction:
\[ \Delta_rH = (+369 \text{ kJ}) + (-483 \text{ kJ}) = -114 \text{ kJ/mol} \]

### Final Answer:
Thus, the standard reaction enthalpy (Δ_rH^@) for the reaction is:
\[ \Delta_rH^@ = -114 \text{ kJ/mol} \]

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