The free energy change for the following reactions are given below
`C_(2)H_(2)(g)+(5)/(2)O_(2)(g)to2CO_(2)(g)+H_(2)O(l),DeltaG^(@)=-1234kJ`
`C(s)+O_(2)(g)toCO_(2)(g),DeltaG^(@)=-394kJ`
`H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),DeltaG^(@)=-273kJ`
What is the standard free energy change for the reaction `H_(2)(g)+2C(s)toC_(2)H_(2)(g)` :-

Options

A. (a)`-209 kJ`
B. `(b)-2259 kJ`
C. (c)`+2259 kJ`
D. (d)`209 kJ`

Answer

To find the standard free energy change for the reaction:

\[ H_2(g) + 2C(s) \rightarrow C_2H_2(g) \]

we will use the given reactions and their respective free energy changes. The reactions provided are:

1. \( C_2H_2(g) + \frac{5}{2}O_2(g) \rightarrow 2CO_2(g) + H_2O(l), \Delta G^\circ = -1234 \, \text{kJ} \)
2. \( C(s) + O_2(g) \rightarrow CO_2(g), \Delta G^\circ = -394 \, \text{kJ} \)
3. \( H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l), \Delta G^\circ = -273 \, \text{kJ} \)

### Step 1: Identify the reactions to manipulate
We need to manipulate the given reactions to derive the desired reaction. We will use reactions 2 and 3, and we will reverse reaction 1.

### Step 2: Reverse Reaction 1
Reversing reaction 1 gives us:

\[ 2CO_2(g) + H_2O(l) \rightarrow C_2H_2(g) + \frac{5}{2}O_2(g), \Delta G^\circ = +1234 \, \text{kJ} \]

### Step 3: Double Reaction 2
We need 2 moles of carbon (C) for our target reaction, so we will double reaction 2:

\[ 2C(s) + O_2(g) \rightarrow 2CO_2(g), \Delta G^\circ = 2 \times (-394) = -788 \, \text{kJ} \]

### Step 4: Use Reaction 3 as is
We will use reaction 3 as it is:

\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l), \Delta G^\circ = -273 \, \text{kJ} \]

### Step 5: Combine the reactions
Now we combine the modified reactions:

1. From the reversed reaction 1:
\[ 2CO_2(g) + H_2O(l) \rightarrow C_2H_2(g) + \frac{5}{2}O_2(g) \]
2. From the doubled reaction 2:
\[ 2C(s) + O_2(g) \rightarrow 2CO_2(g) \]
3. From reaction 3:
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \]

### Step 6: Write the net reaction
When we add these reactions, we will cancel out the common species:

- The \( 2CO_2(g) \) from reaction 2 cancels with \( 2CO_2(g) \) from the reversed reaction 1.
- The \( H_2O(l) \) from reaction 3 cancels with \( H_2O(l) \) from the reversed reaction 1.
- The \( O_2(g) \) terms will also cancel out.

This results in the net reaction:

\[ H_2(g) + 2C(s) \rightarrow C_2H_2(g) \]

### Step 7: Calculate the standard free energy change
The standard free energy change for the net reaction is given by:

\[ \Delta G^\circ_{\text{net}} = \Delta G^\circ_3 + \Delta G^\circ_2 - \Delta G^\circ_1 \]

Substituting the values:

\[ \Delta G^\circ_{\text{net}} = (-273 \, \text{kJ}) + (-788 \, \text{kJ}) + 1234 \, \text{kJ} \]

Calculating this gives:

\[ \Delta G^\circ_{\text{net}} = -273 - 788 + 1234 = 173 \, \text{kJ} \]

### Final Answer
The standard free energy change for the reaction \( H_2(g) + 2C(s) \rightarrow C_2H_2(g) \) is:

\[ \Delta G^\circ = 173 \, \text{kJ} \]