The standard enthalpy of decomposition of the yellow complex `H_(3)NSO_(2)` into `NH_(3)` and `SO_(2)` is `+40 kJ mol^(-1)`. Calculate the standard enthalpy of formation of `H_(3)NSO_(3).DeltaH_(f)^(0)(NH_(3))= -46.17 kJ mol^(-1), DeltaH_(f)^(0)(SO)_(2)= -296.83`

Answer

To calculate the standard enthalpy of formation of \( H_3NSO_3 \), we will follow these steps:

### Step 1: Understand the reaction
The decomposition of the yellow complex \( H_3NSO_2 \) into \( NH_3 \) and \( SO_2 \) can be represented as:
\[ H_3NSO_2 \rightarrow NH_3 + SO_2 \]
The standard enthalpy change for this reaction is given as \( \Delta H_{reaction} = +40 \, \text{kJ/mol} \).

### Step 2: Write the enthalpy change equation
According to Hess's law, the enthalpy change of a reaction can be expressed as:
\[ \Delta H_{reaction} = \Delta H_{products} - \Delta H_{reactants} \]
In this case, we can rearrange the equation to find the enthalpy of formation of \( H_3NSO_2 \):
\[ \Delta H_{reactants} = \Delta H_{products} - \Delta H_{reaction} \]

### Step 3: Substitute known values
We know:
- \( \Delta H_f^0(NH_3) = -46.17 \, \text{kJ/mol} \)
- \( \Delta H_f^0(SO_2) = -296.83 \, \text{kJ/mol} \)
- \( \Delta H_{reaction} = +40 \, \text{kJ/mol} \)

Now, we can calculate \( \Delta H_{products} \):
\[ \Delta H_{products} = \Delta H_f^0(NH_3) + \Delta H_f^0(SO_2) = -46.17 \, \text{kJ/mol} + (-296.83 \, \text{kJ/mol}) = -343 \, \text{kJ/mol} \]

### Step 4: Calculate \( \Delta H_{reactants} \)
Now, substituting into the rearranged equation:
\[ \Delta H_{reactants} = -343 \, \text{kJ/mol} - 40 \, \text{kJ/mol} = -383 \, \text{kJ/mol} \]

### Step 5: Conclusion
Thus, the standard enthalpy of formation of \( H_3NSO_2 \) is:
\[ \Delta H_f^0(H_3NSO_2) = -383 \, \text{kJ/mol} \]

### Step 6: Calculate \( \Delta H_f^0(H_3NSO_3) \)
To find the enthalpy of formation of \( H_3NSO_3 \), we need to consider the reaction:
\[ H_3NSO_2 + O \rightarrow H_3NSO_3 \]
Assuming the enthalpy change for this reaction is \( \Delta H \), we can express it as:
\[ \Delta H_f^0(H_3NSO_3) = \Delta H_f^0(H_3NSO_2) + \Delta H \]
If we assume the enthalpy change for the oxidation of \( H_3NSO_2 \) to \( H_3NSO_3 \) is known or can be calculated, we can find \( \Delta H_f^0(H_3NSO_3) \).