Boron can undergo the following reactions with the given enthalpy changes :
`2B(s)+(3)/(2)O_(2)(g)rarrB_(2)O_(3)(s)," "DeltaH=-1260 " kJ"`
`2B(s)+3H_(2)(g)rarrB_(2)H_(6)(g)," "DeltaH=30 " kJ"`
Assume no other reactions are occuring.
If in a container (operating at constant pressure) which is isolated from the surrounding, mixture of `H_(2)` (gas) and `O_(2)` (gas) are passed over excess of B(s), then calculate the molar ratio `(O_(2) : H_(2))` so that temperature of the container do not change :

Options

A. `15 : 3`
B. `42 : 1`
C. `1 : 42`
D. `1 : 84`

Explanation

To solve the problem, we need to determine the molar ratio of \( O_2 \) to \( H_2 \) such that the temperature of the container remains constant when both gases are passed over excess boron \( B(s) \).

### Step-by-Step Solution:

1. **Identify the Reactions and Their Enthalpy Changes:**
- The first reaction is:
\[ 2B(s) + \frac{3}{2}O_2(g) \rightarrow B_2O_3(s), \quad \Delta H = -1260 \, \text{kJ} \]
- The second reaction is:
\[ 2B(s) + 3H_2(g) \rightarrow B_2H_6(g), \quad \Delta H = 30 \, \text{kJ} \]

2. **Set Up the Energy Balance:**
- For the temperature of the container to remain constant, the heat released by the formation of \( B_2O_3 \) must equal the heat absorbed by the formation of \( B_2H_6 \).
- Therefore, we can write:
\[ q_{\text{released}} + q_{\text{absorbed}} = 0 \]
- This implies:
\[ -\Delta H_{\text{reaction 1}} + \Delta H_{\text{reaction 2}} = 0 \]

3. **Calculate the Moles of \( O_2 \) Required for the Second Reaction:**
- The enthalpy change for the first reaction is \( -1260 \, \text{kJ} \) for the formation of \( B_2O_3 \) from \( 2B \) and \( \frac{3}{2}O_2 \).
- The enthalpy change for the second reaction is \( 30 \, \text{kJ} \) for the formation of \( B_2H_6 \) from \( 2B \) and \( 3H_2 \).

4. **Relate the Moles of \( O_2 \) to the Enthalpy Changes:**
- From the first reaction, \( \frac{3}{2} \) moles of \( O_2 \) release \( 1260 \, \text{kJ} \).
- Therefore, the energy released per mole of \( O_2 \) is:
\[ \frac{1260 \, \text{kJ}}{\frac{3}{2}} = 840 \, \text{kJ/mole} \]

5. **Determine the Moles of \( O_2 \) Needed for \( 30 \, \text{kJ} \):**
- To find out how many moles of \( O_2 \) are required to provide \( 30 \, \text{kJ} \):
\[ \text{Moles of } O_2 = \frac{30 \, \text{kJ}}{840 \, \text{kJ/mole}} = \frac{1}{28} \, \text{moles of } O_2 \]

6. **Determine the Moles of \( H_2 \) Required:**
- From the second reaction, \( 3 \, \text{moles of } H_2 \) are required for the formation of \( B_2H_6 \).
- Therefore, the molar ratio of \( O_2 \) to \( H_2 \) can be calculated as follows:
\[ \text{Moles of } H_2 = 3 \, \text{moles} \]

7. **Calculate the Molar Ratio \( O_2:H_2 \):**
- The molar ratio of \( O_2 \) to \( H_2 \) is:
\[ \text{Ratio} = \frac{\text{Moles of } O_2}{\text{Moles of } H_2} = \frac{\frac{1}{28}}{3} = \frac{1}{84} \]

### Final Answer:
The molar ratio of \( O_2 : H_2 \) is \( 1 : 84 \).