For a reversible exothermic reaction
`K_(c) lt K_(p) & DeltaH=-100kJ` the reverse reaction is favoured if `:`

Options

A. Both `P & T` are reduced
B. `P` increased `& T` decreased
C. Both `P & T` are increased
D. `P` decreased `& T` increased

Explanation

To determine the conditions under which the reverse reaction of a given exothermic reaction is favored, we can follow these steps:

### Step 1: Understand the Reaction
Given that the reaction is exothermic, we know that it releases heat. The enthalpy change (ΔH) for the forward reaction is negative (-100 kJ).

### Step 2: Analyze the Relationship between Kc and Kp
The relationship between Kc and Kp is given by the equation:
\[ K_p = K_c R T^{\Delta N_g} \]
where \( \Delta N_g \) is the change in the number of moles of gaseous species (moles of products - moles of reactants).

### Step 3: Determine the Implications of Kc < Kp
Since it is given that \( K_c < K_p \), this implies that \( \Delta N_g \) must be positive. Therefore, the reaction produces more moles of gas on the product side compared to the reactant side.

### Step 4: Conditions Favoring the Reverse Reaction
1. **Pressure**:
- For reactions where the number of moles increases in the forward direction (as indicated by \( \Delta N_g > 0 \)), increasing the pressure will favor the reverse reaction (which has fewer moles).

2. **Temperature**:
- Since the forward reaction is exothermic, it is favored at low temperatures. Conversely, the reverse reaction (which is endothermic) will be favored at high temperatures.

### Step 5: Conclusion
To favor the reverse reaction, we need to increase both the pressure and the temperature. Thus, the correct answer is that the reverse reaction is favored if both pressure and temperature are increased.

### Final Answer:
The reverse reaction is favored if both pressure and temperature are increased.

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