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Determine the heat of transformaiton of...

Determine the heat of transformaiton of `C_(("dimamond"))rarrC_(("graphite"))` form the following data:
i. `C_(("diamond"))+O_(2)(g) rarr CO_(2)(g), DeltaH^(Theta) =- 94.5 kcal`
ii. `C_(("graphite")) +O_(2)(g) rarr CO_(2)(g),DeltaH^(Theta) =- 94.0 kcal`

Text Solution

Verified by Experts

Substracting equation (ii) from equation (i), the required equation is obtained.
`C_(("diamone")) rarr C(g)`
`Delta_("trans")H^(Theta) =- 94.5-(-94.0) =- 94.5 + 94.0`
`=- 0.5 kcal`
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Calculate the heat of formation of acetic acid form the following date: a. CH_(3)COOH(l) +2O_(2)(g) rarr 2CO_(2)(g) +2H_(2)O(l)DeltaH^(Theta) =- 200.0kcal b. C(s) +O_(2) (g) rarr CO_(2)(g), DeltaH^(Theta) =- 94.0 kcal c. H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(l), DeltaH^(Theta) =- 68.0 kcal

Knowledge Check

  • The value of DeltaH for the conversion of C (diamond) to C (graphite) when the following reactions are given C(diamond) + O_(2)(g) to CO_(2)(g), DeltaH=-94.5 kcal C(graphite) + O_(2)(g) to CO_(2)(g), DeltaH=-94.0 kcal is:

    A
    `-188.5` kcal
    B
    `+188.5` kcal
    C
    `+0.5` kcal
    D
    `-0.5` kcal

  • C_(("graphite"))+O_(2)(g)rarrCO_(2)(g), DeltaH=-94.05 kcal mol^(-1) C_(("diamond"))+O_(2)(g)rarrCO_(2)(g), DeltaH=-94.50 kcal mol^(-1) therefore

    A
    `C_(("graphite"))rarrC_(("diamond")), DeltaH_(298 K)^(@)=-450 cal mol^(-1)`
    B
    `C_(("diamond"))rarrC_(("graphite")), DeltaH_(298 K)^(@)=+450 cal mol^(-1)`
    C
    Graphite is the stabler allotrope
    D
    Diamond is harder than graphite

  • In the reaction C_((s)) +O_(2(g)) rarr CO_(2(g)), Delta H =-94.3 Kcal

    A
    `H_(P) gt H_(R)`
    B
    `H_(P) lt H_(R)`
    C
    `H_(P) = H_(R)`
    D
    `H_(P)=2H_(R)`

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