In a reaction, all reactant and products are liquid, then

Options

A. `Delta H gt Delta E`
B. `Delta H lt Delta E`
C. `Delta H = Delta E`
D. Can't predicted

Answer

To solve the question regarding the relationship between ΔH (change in enthalpy) and ΔE (change in internal energy) when all reactants and products are in liquid form, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**: We start with the equation that relates the change in enthalpy (ΔH) to the change in internal energy (ΔE):
\[ \Delta H = \Delta E + \Delta N_g RT \]
where ΔN_g is the change in the number of moles of gas during the reaction, R is the universal gas constant, and T is the temperature in Kelvin.

2. **Identify the State of Reactants and Products**: In this case, all reactants and products are in the liquid state. This is crucial because it affects the value of ΔN_g.

3. **Determine ΔN_g**: Since there are no gaseous reactants or products (all are liquids), the change in the number of moles of gas (ΔN_g) is zero:
\[ \Delta N_g = 0 \]

4. **Substitute ΔN_g into the Equation**: Plugging ΔN_g = 0 into the relationship gives:
\[ \Delta H = \Delta E + 0 \cdot RT \]
This simplifies to:
\[ \Delta H = \Delta E \]

5. **Conclusion**: Therefore, when all reactants and products are in the liquid state, the change in enthalpy (ΔH) is equal to the change in internal energy (ΔE).

### Final Answer:
\[ \Delta H = \Delta E \]