The values of `DeltaH` and `DeltaS` for the reaction,
`C_("graphite")+CO_(2)(g)rarr2CO(g)`
are `170KJ` and `170JK^(-1)` respectively. This reaction will be spontaneous at

Options

A. 710 K
B. 910 K
C. 1110 K
D. 510 K

Explanation

To determine the temperature at which the reaction \( C_{\text{(graphite)}} + CO_2(g) \rightarrow 2CO(g) \) becomes spontaneous, we can use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

For a reaction to be spontaneous, \(\Delta G\) must be negative. Therefore, we need to find the temperature \(T\) at which:

\[ \Delta H - T \Delta S < 0 \]

This can be rearranged to:

\[ T \Delta S > \Delta H \]

### Step 1: Convert \(\Delta H\) to Joules
Given:
- \(\Delta H = 170 \, \text{kJ} = 170 \times 10^3 \, \text{J}\)
- \(\Delta S = 170 \, \text{J/K}\)

### Step 2: Set up the inequality
We need to find \(T\) such that:

\[ T \cdot 170 \, \text{J/K} > 170 \times 10^3 \, \text{J} \]

### Step 3: Solve for \(T\)
Rearranging the inequality gives:

\[ T > \frac{170 \times 10^3 \, \text{J}}{170 \, \text{J/K}} \]

### Step 4: Simplify the expression
The \(170\) in the numerator and denominator cancels out:

\[ T > 1000 \, \text{K} \]

### Conclusion
The reaction will be spontaneous at temperatures greater than \(1000 \, \text{K}\).