The value of `DeltaH` and `DeltaS` for the reaction
`C_"(graphite)"+O_(2(g))to CO_(2(g))` are -100KJ and `-100JK^(-1)`
respectively. The reaction will be spontaneous at :-

Options

A. 1000 K
B. 900 K
C. 1100 K
D. At any temperture

Explanation

To determine the temperature at which the reaction \( C_{\text{(graphite)}} + O_{2(g)} \rightarrow CO_{2(g)} \) is spontaneous, we need to analyze the given values of enthalpy change (\( \Delta H \)) and entropy change (\( \Delta S \)).

### Step-by-Step Solution:

1. **Identify the given values:**
- \( \Delta H = -100 \, \text{kJ} = -100,000 \, \text{J} \) (since 1 kJ = 1000 J)
- \( \Delta S = -100 \, \text{J/K} \)

2. **Use the Gibbs Free Energy equation:**
The Gibbs Free Energy (\( \Delta G \)) is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]
For the reaction to be spontaneous, \( \Delta G \) must be less than zero:
\[ \Delta G < 0 \]

3. **Set up the inequality:**
To find the temperature at which the reaction becomes spontaneous, we set up the inequality:
\[ \Delta H - T \Delta S < 0 \]

4. **Rearrange the inequality:**
Rearranging gives:
\[ T \Delta S > \Delta H \]
or
\[ T > \frac{\Delta H}{\Delta S} \]

5. **Substitute the values:**
Substitute the values of \( \Delta H \) and \( \Delta S \):
\[ T > \frac{-100,000 \, \text{J}}{-100 \, \text{J/K}} \]

6. **Calculate the temperature:**
\[ T > 1000 \, \text{K} \]

7. **Conclusion:**
The reaction will be spontaneous at temperatures less than 1000 K. Therefore, if we have options, we choose the highest temperature that is less than 1000 K. If the options include 900 K, that would be the answer.

### Final Answer:
The reaction will be spontaneous at temperatures less than 1000 K, specifically at 900 K.

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