The heats of combustion of `C_(x)H_(y)`, carbon and hydrogen are a, b and c cal respectively. The heat of formation of `C_(x)H_(y)`, will be:

Options

A. `-(xb+(yc)/(2)-(a)/(2))" cal"`
B. `-(xb+(yc)/(2)-a)" cal"`
C. `(xb-(yc)/(2)+(a)/(2))" cal"`
D. `(xb-(yc)/(2)-(a)/(2))" cal"`

Answer

To find the heat of formation of \( C_xH_y \) using the heats of combustion of \( C_xH_y \), carbon, and hydrogen, we can follow these steps:

### Step 1: Write the formation reaction
The formation reaction for \( C_xH_y \) can be represented as:
\[ x \text{ moles of } C (s) + \frac{y}{2} \text{ moles of } H_2 (g) \rightarrow C_xH_y (g) \]

### Step 2: Understand the concept of heat of combustion
The heat of combustion (\( \Delta H_c \)) is the heat released when a substance is burned in excess oxygen. The heats of combustion for the substances involved are given as:
- \( A \) for \( C_xH_y \)
- \( B \) for carbon
- \( C \) for hydrogen

### Step 3: Apply Hess's Law
According to Hess's Law, the heat of formation (\( \Delta H_f \)) can be calculated using the heats of combustion:
\[ \Delta H_f = \Delta H_c (\text{reactants}) - \Delta H_c (\text{products}) \]

### Step 4: Set up the equation
For the combustion of \( C_xH_y \):
\[ \Delta H_c (C_xH_y) = A \]
For the combustion of \( x \) moles of carbon:
\[ \Delta H_c (C) = xB \]
For the combustion of \( \frac{y}{2} \) moles of hydrogen:
\[ \Delta H_c (H) = \frac{y}{2}C \]

### Step 5: Substitute into the equation
Now substituting these values into the equation from Step 3:
\[ \Delta H_f = A - (xB + \frac{y}{2}C) \]

### Step 6: Rearranging the equation
Rearranging gives us:
\[ \Delta H_f = A - xB - \frac{y}{2}C \]

### Final Answer
Thus, the heat of formation of \( C_xH_y \) is:
\[ \Delta H_f = A - xB - \frac{y}{2}C \]

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