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The enthalpy change for chemical reactio...

The enthalpy change for chemical reaction is denoted aas `DeltaH^(Theta)` and `DeltaH^(Theta) = H_(P)^(Theta) - H_(R)^(Theta)`. The relation between enthalpy and internal enegry is expressed by equation:
`DeltaH = DeltaU +DeltanRT`
where `DeltaU =` change in internal enegry `Deltan =` change in number of moles, `R =` gas constant.
For a reaction, `2X(s) +2Y(s) rarr 2C(l) +D(g), DeltaH` at `27^(@)C` is `-28 kcal mol^(-1). DeltaU` is .....`kcal mol^(-1)`

A

`-25.5`

B

`+25.5`

C

`-28.6`

D

`28.4`

Text Solution

Verified by Experts

`DeltaH = DeltaU +DeltanRT`
`:. DeltaU = DeltaH - DeltanRT`
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The enthalpy change for chemical reaction is denoted aas DeltaH^(Theta) and DeltaH^(Theta) = H_(P)^(Theta) - H_(R)^(Theta) . The relation between enthalpy and internal enegry is expressed by equation: DeltaH = DeltaU +DeltanRT where DeltaU = change in internal enegry Deltan = change in number of moles, R = gas constant. Enthalpy of the system is given as

The enthalpy change for chemical reaction is denoted aas DeltaH^(Theta) and DeltaH^(Theta) = H_(P)^(Theta) - H_(R)^(Theta) . The relation between enthalpy and internal enegry is expressed by equation: DeltaH = DeltaU +DeltanRT where DeltaU = change in internal enegry Deltan = change in number of moles, R = gas constant. Which of the following equations corresponds to the definition of enthalpy of formation at 298K ?

The enthalpy change for chemical reaction is denoted as DeltaH^(Theta) and DeltaH^(Theta) = H_(P)^(Theta) - H_(R)^(Theta) . The relation between enthalpy and internal enegry is expressed by equation: DeltaH = DeltaU +DeltanRT where DeltaU = change in internal enegry Deltan = change in number of moles, R = gas constant. For the change, C_("diamond") rarr C_("graphite"), DeltaH =- 1.89 kJ , if 6g of diamond and 6g of graphite are seperately burnt to yield CO_(2) the heat liberated in first case is

The enthalpy change for chemical reaction is denoted aas DeltaH^(Theta) and DeltaH^(Theta) = H_(P)^(Theta) - H_(R)^(Theta) . The relation between enthalpy and internal enegry is expressed by equation: DeltaH = DeltaU +DeltanRT where DeltaU = change in internal enegry Deltan = change in number of moles, R = gas constant. H_(2)(g) +((1)/(2))O_(2)(g) = H_(2)O(l), DeltaH_(298K) = - 68.00kcal Heat of voporisation of water at 1 atm and 25^(@)C is 10.00 kcal . The standard heat of formation (in kcal) of 1 amol vapour a 25^(@)C is

The relation between DeltaU and DeltaH :

Enthalpy of reaction DeltaH is expressed as

For gaseous reactions, if DeltaH is the change in enthalpy and DeltaU that in internal energy then:

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