`2Zn + O_(2) rightarrow 2ZnO, Delta G^(@) = -616 J`
`2Zn + S_(2) rightarrow 2ZnS, Delta G^(@) = -293 J`
` S_(2) + 2O_(2) rightarrow 2SO_(2), Delta G^(@) = -408 J`
`Delta G^(@)` for the following reaction
`2ZnS + 3O_(2) rightarrow 2ZnO+ 2 SO_(2) ` is

Options

A. `-713J`
B. `-1317J`
C. `-501J`
D. `+731J`

Answer

To find the ΔG for the reaction:

\[ 2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2 \]

we will use the given reactions and their corresponding ΔG values.

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

1. \( 2Zn + O_2 \rightarrow 2ZnO, \quad \Delta G^(@) = -616 \, \text{J} \)
2. \( 2Zn + S_2 \rightarrow 2ZnS, \quad \Delta G^(@) = -293 \, \text{J} \)
3. \( S_2 + 2O_2 \rightarrow 2SO_2, \quad \Delta G^(@) = -408 \, \text{J} \)

### Step 2: Reverse the second reaction.

Since we need \( ZnS \) in the reactants, we will reverse the second reaction:

\[ 2ZnS \rightarrow 2Zn + S_2 \]

Now, the ΔG for this reversed reaction will be:

\[ \Delta G^(@) = +293 \, \text{J} \]

### Step 3: Write the first reaction as it is.

The first reaction can be used directly since it already has \( ZnO \) in the products:

\[ 2Zn + O_2 \rightarrow 2ZnO \]

With ΔG:

\[ \Delta G^(@) = -616 \, \text{J} \]

### Step 4: Write the third reaction as it is.

The third reaction can also be used directly since it has \( SO_2 \) in the products:

\[ S_2 + 2O_2 \rightarrow 2SO_2 \]

With ΔG:

\[ \Delta G^(@) = -408 \, \text{J} \]

### Step 5: Combine the reactions.

Now we will combine the three reactions:

1. From the reversed second reaction:
\[ 2ZnS \rightarrow 2Zn + S_2 \quad (\Delta G = +293 \, \text{J}) \]

2. From the first reaction:
\[ 2Zn + O_2 \rightarrow 2ZnO \quad (\Delta G = -616 \, \text{J}) \]

3. From the third reaction:
\[ S_2 + 2O_2 \rightarrow 2SO_2 \quad (\Delta G = -408 \, \text{J}) \]

When we add these reactions, we will cancel out the \( 2Zn \) and \( S_2 \):

- The \( 2Zn \) from the first reaction cancels with the \( 2Zn \) produced from the reversed second reaction.
- The \( S_2 \) from the third reaction cancels with the \( S_2 \) produced from the reversed second reaction.

The overall reaction becomes:

\[ 2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2 \]

### Step 6: Calculate the overall ΔG.

Now we can calculate the overall ΔG by adding the ΔG values of the individual reactions:

\[ \Delta G = (+293) + (-616) + (-408) \]

Calculating this gives:

\[ \Delta G = 293 - 616 - 408 = -731 \, \text{J} \]

### Final Answer:

The ΔG for the reaction \( 2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2 \) is:

\[ \Delta G = -731 \, \text{J} \]