The molar heat capacities at constant pressure (assumed constant with respect to temperature ) of A,B and C are in ratio 3:1.5:2.0. the enthyalpy change for the exothermic reaction `A+ 2B to 3C ` at 300k and 310 K is `Delta H_(300)` and `DeltaH_(310)` respectively then:

Options

A. `DeltaH_(300)gtDeltaH_(310)`
B. `DeltaH_(300)ltDeltaH_(310)`
C. `DeltaH_(300)=DeltaH_(310)`
D. if `T_(2)gtT_(1)" then "DeltaH_(310)gtDeltaH_(300)" and if "T_(2)ltT_(1)" then "DeltaH_(310)ltDeltaH_(300)`

Explanation

To solve the problem, we need to analyze the given information about the molar heat capacities of substances A, B, and C, and how they relate to the enthalpy change of the reaction at different temperatures.

### Step-by-Step Solution:

1. **Identify the Molar Heat Capacities**:
The molar heat capacities at constant pressure for A, B, and C are given in the ratio 3:1.5:2.0. We can express these as:
- \( C_{pA} = 3x \)
- \( C_{pB} = 1.5x \)
- \( C_{pC} = 2.0x \)
where \( x \) is a constant.

2. **Write the Reaction**:
The reaction is given as:
\[ A + 2B \rightarrow 3C \]

3. **Calculate the Change in Enthalpy (\( \Delta H \))**:
The change in enthalpy for the reaction can be expressed using the heat capacities:
\[ \Delta H = \sum \nu_i C_{p,i} \Delta T \]
where \( \nu_i \) are the stoichiometric coefficients of the reactants and products.

4. **Calculate \( \Delta C_p \) for the Reaction**:
Using the stoichiometric coefficients:
- For products (3C): \( 3 \times C_{pC} = 3 \times 2.0x = 6.0x \)
- For reactants (A + 2B): \( 1 \times C_{pA} + 2 \times C_{pB} = 3x + 2 \times 1.5x = 3x + 3x = 6.0x \)

Therefore, the change in heat capacity (\( \Delta C_p \)) for the reaction is:
\[ \Delta C_p = 6.0x - 6.0x = 0 \]

5. **Relate Enthalpy Changes at Different Temperatures**:
The change in enthalpy at two different temperatures can be expressed as:
\[ \Delta H_{310} - \Delta H_{300} = \Delta C_p \times (T_2 - T_1) \]
Substituting \( \Delta C_p = 0 \):
\[ \Delta H_{310} - \Delta H_{300} = 0 \times (310 - 300) = 0 \]

6. **Conclusion**:
This implies:
\[ \Delta H_{310} = \Delta H_{300} \]
Therefore, the enthalpy change for the reaction does not depend on the temperature in this case.

### Final Answer:
The enthalpy change at 310 K is equal to the enthalpy change at 300 K, i.e., \( \Delta H_{310} = \Delta H_{300} \).