For the reaction
`CH_4(g) + 2O_2(g) hArr CO_2(g) + 2H_2O(l) , Delta_rH=-170.8 "kJ mol"^(-1)`
Which of the following statement is not true?

Options

A. The equilibrium constant for the reaction is given by `K_p=(pCO_2)/(pCH_4 xxpO_2)`
B. Addition of `CH_4(g)` or `O_2` at equilibrium will cause a shift to the right
C. The reaction is exothermic
D. At equilibrium, the concentrations of `CO_2(g)` and `H_2O(l)` are not equal .

Explanation

To determine which statement is not true regarding the reaction:

\[ CH_4(g) + 2O_2(g) \rightleftharpoons CO_2(g) + 2H_2O(l), \Delta_rH = -170.8 \text{ kJ mol}^{-1} \]

we will analyze each option step by step.

### Step 1: Understanding the Reaction and Kp
The reaction involves methane (CH₄) and oxygen (O₂) reacting to form carbon dioxide (CO₂) and water (H₂O). The enthalpy change (\(\Delta_rH\)) is negative, indicating that the reaction is exothermic.

**Hint:** Remember that Kp is calculated using only gaseous species. Liquids do not appear in the equilibrium expression.

### Step 2: Writing the Expression for Kp
The equilibrium constant \(K_p\) for this reaction can be expressed as:

\[ K_p = \frac{P_{CO_2}}{P_{CH_4} \cdot P_{O_2}^2} \]

Since water (H₂O) is in the liquid state, it does not appear in the expression for \(K_p\).

**Hint:** Identify which species are gases and which are liquids when writing equilibrium expressions.

### Step 3: Evaluating Option A
If Option A states something incorrect about the calculation of \(K_p\), it might suggest including the liquid water in the expression or misrepresenting the relationship between the partial pressures.

**Hint:** Check if the option incorrectly includes liquid species in the equilibrium expression.

### Step 4: Analyzing the Effect of Adding Reactants (Option B)
When more CH₄ or O₂ is added to the system at equilibrium, according to Le Chatelier's principle, the equilibrium will shift to the right to produce more products (CO₂ and H₂O).

**Hint:** Recall Le Chatelier's principle regarding changes in concentration.

### Step 5: Evaluating the Exothermic Nature (Option C)
The reaction is exothermic, as indicated by the negative \(\Delta_rH\). This means that heat is released during the reaction.

**Hint:** Remember the definitions of exothermic and endothermic reactions.

### Step 6: Concentration of Products and Reactants at Equilibrium (Option D)
At equilibrium, the concentrations of CO₂ and H₂O will not be equal due to the stoichiometry of the reaction (1:2 ratio). Therefore, if an option states that their concentrations are equal, it is incorrect.

**Hint:** Consider the stoichiometric coefficients when comparing concentrations of products.

### Conclusion
After analyzing all options, we find that:

- Option A incorrectly represents the calculation of \(K_p\) by including liquid water.
- Options B and C are correct based on Le Chatelier's principle and the nature of the reaction.
- Option D correctly states that the concentrations of CO₂ and H₂O are not equal.

Thus, the statement that is **not true** is **Option A**.

### Final Answer
**The statement that is not true is Option A.**