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

Options

A. `DeltaH_(300) gt DeltaH_(310)`
B. `DeltaH_(300) lt DeltaH_(310)`
C. `DeltaH_(300) = DeltaH_(310)`
D. if `T_(2) gt T_(1)` then `DeltaH_(310) gt DeltaH_(300)` and if `T_(2) lt T_(1)` then `DeltaH_(310) lt DeltaH_(300)`

Explanation

To solve the problem, we need to analyze the given information about the molar heat capacities and how they relate to the enthalpy changes at two different temperatures for the reaction:

**Reaction:**
\[ A + 2B \rightarrow 3C \]

**Step 1: Understand the Molar Heat Capacities**
The molar heat capacities at constant pressure for A, B, and C are given in the ratio of 3:1.5:2.0. To simplify calculations, we can convert this ratio into integers:
- Let the molar heat capacities be \( C_A = 3x \), \( C_B = 1.5x \), and \( C_C = 2x \).
- To express them in integers, multiply each term by 2:
- \( C_A = 6 \), \( C_B = 3 \), \( C_C = 4 \).

**Step 2: Calculate the Change in Heat Capacity (ΔCp) for the Reaction**
The change in heat capacity for the reaction can be calculated using the formula:
\[ \Delta C_p = C_{products} - C_{reactants} \]
For the reaction:
- Products: \( 3C \) contributes \( 3 \times C_C = 3 \times 4 = 12 \)
- Reactants: \( A + 2B \) contributes \( C_A + 2C_B = 6 + 2 \times 3 = 6 + 6 = 12 \)

Thus,
\[ \Delta C_p = 12 - 12 = 0 \]

**Step 3: Relate Enthalpy Changes at Different Temperatures**
The relationship between the enthalpy changes at two different temperatures can be expressed as:
\[ \Delta H_{310} = \Delta H_{300} + \Delta V \Delta C_p \]
Where \( \Delta V \) is the change in volume (which we assume to be constant for this reaction). Since we found that \( \Delta C_p = 0 \):
\[ \Delta H_{310} = \Delta H_{300} + 0 \]

**Step 4: Conclusion**
From the above relationship, we can conclude that:
\[ \Delta H_{310} = \Delta H_{300} \]

Thus, the enthalpy change at 310 K is equal to the enthalpy change at 300 K.

**Final Answer:**
\[ \Delta H_{310} = \Delta H_{300} \]

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