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Will the heat released be same or differ...

Will the heat released be same or different in the following two reactions:
(i)`H_(2)(g) + 1/2 O_(2)(g) to H_(2)O(l)` (ii) `H_(2)(g) + 1/2 O_(2)(g) to H_(2)O(g)`
Explain.

Text Solution

AI Generated Solution

To determine whether the heat released in the two reactions is the same or different, we need to analyze the processes involved in each reaction. ### Step-by-Step Solution: 1. **Identify the Reactions**: - Reaction (i): \( H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \) - Reaction (ii): \( H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g) \) ...
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In the reaction 2H_(2)(g) + O_(2)(g) to 2H_(2)O(l), DeltaH = -x kJ

Calculate the entropy changes for the following reactions : (i) 2CO(g) + O_(2)(g) to 2CO_(2)(g) (ii) 2H_(2)(g) + O_(2)(g) to 2H_(2)O(l) Entropies of different compounds are : CO(g) = 197.6 J K^(-1) "mol"^(-1), O_(2)(g) = 205.03 J K^(-1) "mol"^(-1) CO_(2)(g) = 213.6 JK^(-1) "mol"^(-1), H_(2)(g) = 130.6 J K^(-1) "mol"^(-1) H__(2)O(l) = 69.96 J K^(-1) "mol"^(-1)

Knowledge Check

  • The enthalpy of reaction, H_(2)(g)+(1)/(2)O_(2)(g)to H_(2)O(g) is DeltaH_(1) and that of H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l) is DeltaH_(2) . Then

    A
    `DeltaH_(1) lt DeltaH_(2)`
    B
    `DeltaH_(1)+DeltaH_(2)=0`
    C
    `DeltaH_(1) gt DeltaH_(2)`
    D
    `DeltaH_(1)=DeltaH_(2)`

  • The enthalpy change for the reaction- H_2(g) + (1)/(2) O_2(g) to H_2O (g) is called

    A
    Enthalpy of formation of water
    B
    Enthalpy of solution
    C
    Enthalpy of vaporisation of water
    D
    None of these

  • For the two equations given below: H_(2)(g)+1//2O_(2)(g)toH_(2)O(l)+x_(1)kJ H_(2)(g)+1//2O_(2)(g)toH_(2)O(g)+x_(2)kJ Select the correct answer:

    A
    `x_(1) gt x_2`
    B
    `x_(2) gt x_(1)`
    C
    `x_(1)=x_(2)`
    D
    `x_(1)+x_(2)=0`

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    The enthalpy of the reaction H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(g) is DeltaH_(1) and that of H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l) is DeltaH_(2) . Then

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