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 find the molar ratio of \( O_2 \) to \( H_2 \) such that the temperature of the container remains constant. This means that the heat released by the reaction of boron with oxygen must equal the heat absorbed by the reaction of boron with hydrogen.

### Step-by-Step Solution:

1. **Identify the reactions and their enthalpy changes:**
- Reaction 1:
\[ 2B(s) + \frac{3}{2}O_2(g) \rightarrow B_2O_3(s), \quad \Delta H = -1260 \text{ kJ} \]
- Reaction 2:
\[ 2B(s) + 3H_2(g) \rightarrow B_2H_6(g), \quad \Delta H = +30 \text{ kJ} \]

2. **Set up the heat balance equation:**
Since the temperature of the container should not change, the heat released in Reaction 1 must equal the heat absorbed in Reaction 2. Therefore, we can write:
\[ \text{Heat released by Reaction 1} = \text{Heat absorbed by Reaction 2} \]
This can be expressed as:
\[ -\Delta H_1 = \Delta H_2 \]

3. **Calculate the heat released per mole of \( O_2 \):**
From Reaction 1, we see that for every 1 mole of \( B_2O_3 \) produced, \( \frac{3}{2} \) moles of \( O_2 \) are consumed, releasing 1260 kJ of heat. Thus, the heat released per mole of \( O_2 \) is:
\[ \text{Heat released per mole of } O_2 = \frac{-1260 \text{ kJ}}{\frac{3}{2}} = -840 \text{ kJ} \]

4. **Calculate the heat absorbed per mole of \( H_2 \):**
From Reaction 2, we see that for every 3 moles of \( H_2 \) consumed, 30 kJ of heat is absorbed. Thus, the heat absorbed per mole of \( H_2 \) is:
\[ \text{Heat absorbed per mole of } H_2 = \frac{30 \text{ kJ}}{3} = 10 \text{ kJ} \]

5. **Establish the relationship between \( O_2 \) and \( H_2 \):**
Let \( x \) be the number of moles of \( O_2 \) reacted. The heat released by \( x \) moles of \( O_2 \) is:
\[ \text{Heat released} = x \times (-840 \text{ kJ}) \]
Let \( y \) be the number of moles of \( H_2 \) reacted. The heat absorbed by \( y \) moles of \( H_2 \) is:
\[ \text{Heat absorbed} = y \times 10 \text{ kJ} \]

Setting the heat released equal to the heat absorbed gives:
\[ -840x = 10y \]

6. **Express the molar ratio \( O_2 : H_2 \):**
Rearranging the equation gives:
\[ \frac{y}{x} = \frac{-840}{10} = -84 \]
Since we are looking for the ratio of \( O_2 \) to \( H_2 \):
\[ \frac{x}{y} = \frac{1}{84} \]
Therefore, the molar ratio \( O_2 : H_2 \) is:
\[ O_2 : H_2 = 1 : 84 \]

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