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C(2)H(6)(g) + 3.5O(2)(g) rarr 2CO(2)(g) ...

`C_(2)H_(6)(g) + 3.5O_(2)(g) rarr 2CO_(2)(g) + 3H_(2)O(g)`
`DeltaS_("vap") (H_(2)O,l) = "x"_(1)calK^(-1)` (boiling point `=T_(1)`)
`DeltaH_(f)(H_(2)O,l) = "x"_(2)`
`DeltaH_(f)(CO_(2)) = "x"_(3)`
`DeltaH_(f)(C_(2)H_(6)) = "x"_(4)`
Hence , `DeltaH` for the reaction is-

A

`2"x"_(3) + 3"x"_(2)-"x"_(4)`

B

`2"x"_(3) + 3"x"_(2)-"x"_(4) + 3"x"_(1)T_(1)`

C

`2"x"_(3) + 3"x"_(2)-"x"_(4) - 3"x"_(1)T_(1)`

D

`"x"_(1)T_(1) + "X"_(2) + "X"_(3) - "x"_(4)`

Text Solution

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The correct Answer is:
To find the enthalpy change (ΔH) for the given combustion reaction of ethane (C₂H₆), we will use the standard enthalpy of formation (ΔH_f) values for the reactants and products involved in the reaction. The reaction is: \[ C_{2}H_{6}(g) + 3.5O_{2}(g) \rightarrow 2CO_{2}(g) + 3H_{2}O(g) \] ### Step-by-Step Solution: 1. **Identify the Enthalpy of Formation Values**: - ΔH_f (H₂O, l) = x₂ - ΔH_f (CO₂) = x₃ - ΔH_f (C₂H₆) = x₄ - ΔS_vap (H₂O, l) = x₁ (cal/K) at boiling point T₁ 2. **Write the Formula for ΔH of the Reaction**: The enthalpy change for the reaction can be calculated using the formula: \[ \Delta H = \sum \Delta H_f \text{(products)} - \sum \Delta H_f \text{(reactants)} \] 3. **Calculate the Enthalpy for Products**: - For the products: - 2 moles of CO₂ contribute: \( 2 \times x₃ \) - 3 moles of H₂O contribute: \( 3 \times x₂ \) - Therefore, the total enthalpy for products is: \[ \Delta H_{products} = 2x₃ + 3x₂ \] 4. **Calculate the Enthalpy for Reactants**: - For the reactants: - 1 mole of C₂H₆ contributes: \( x₄ \) - The enthalpy of O₂ is zero since it is in its standard state. - Therefore, the total enthalpy for reactants is: \[ \Delta H_{reactants} = x₄ \] 5. **Include the Enthalpy of Vaporization**: - Since water is produced as a gas, we need to account for the vaporization of water. The enthalpy change due to vaporization is given by: \[ 3 \times \Delta S_{vap} \times T_1 = 3x₁T₁ \] 6. **Combine the Results**: Now, substituting the values into the ΔH formula: \[ \Delta H = (2x₃ + 3x₂ + 3x₁T₁) - x₄ \] 7. **Final Expression for ΔH**: Thus, the final expression for ΔH of the reaction is: \[ \Delta H = 2x₃ + 3x₂ - x₄ + 3x₁T₁ \] ### Conclusion: The correct option that matches our derived expression is: \[ \Delta H = 2x₃ + 3x₂ - x₄ + 3x₁T₁ \] This corresponds to option B.
To find the enthalpy change (ΔH) for the given combustion reaction of ethane (C₂H₆), we will use the standard enthalpy of formation (ΔH_f) values for the reactants and products involved in the reaction. The reaction is: \[ C_{2}H_{6}(g) + 3.5O_{2}(g) \rightarrow 2CO_{2}(g) + 3H_{2}O(g) \] ### Step-by-Step Solution: 1. **Identify the Enthalpy of Formation Values**: - ΔH_f (H₂O, l) = x₂ ...
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