Question

Enthalpy change for the reaction $ 4H(g)\to 2{{H}_{2}}(g) $ is $ -869.6kJ $ . The dissociation energy of $ H-H~ $ bond is?

Answer 1

Hint : When a process occurs with pressure kept constant and when the heat is either absorbed or released, then this heat will be equal to the enthalpy change. Enthalpy is called heat content also.. Carbon and Hydrogen are in their elementary states so their enthalpies will be zero. Bond dissociation energy measures the strength of bond.

Complete Step By Step Answer:
It is derived from the Greek meaning called ‘warming’. Entropy and enthalpy are two different terms. It is used in various chemical processes that occur. Or we can say that bond dissociation energy is the energy required to dissociate the bond.
Enthalpy $ \left( H \right) $ is the sum of the internal energy $ \left( U \right) $ and the product of pressure $ \left( P \right) $ and volume $ (V) $ . Enthalpy $ \left( H \right) $ can be written as,
Given balanced equation is , Since Carbon and Hydrogen in this reaction are in their elementary state so their enthalpies will be $ 0 $ and enthalpy for the reaction is So as enthalpies of carbon and hydrogen are zero therefore enthalpy o
 $ \mathbf{H}\text{ }=\text{ }\mathbf{U}\text{ }+\text{ }\mathbf{pV} $
Where, $ H= $ Enthalpy of the system
 $ U\text{ }= $ Internal energy of the system
 $ p\text{ }= $ Pressure of the system
 $ V\text{ }= $ Volume of the system
Change in enthalpy can be measured but not enthalpy directly. It is the measurement of heat added or lost by the system. It is entirely dependent on the state functions $ T,\text{ }p\text{ }and\text{ }U $ Enthalpy can also be written as: $ \Delta H=\Delta U+\Delta PV $
At constant temperature, for the process heat flow(q) is equal to the change in enthalpy, this is represented as: $ \Delta H=q $
The term enthalpy can be expressed as : $ H= $ Energy/Mass
These reactions help to understand how the change in pressure affects the enthalpy.
Thus for the situation above: $ 4H(g)\to 2{{H}_{2}}(g);\,\,\Delta H=~-869.6kJ $
It can be rewritten as : $ 2{{H}_{2}}(g)\to 4H(g);\,\Delta H=869.6kJ $
From the above two reactions we get; the enthalpy divided by $ 2. $
 $ {{H}_{2}}(g)\to 2H(g);\Delta H=869.6/2=434.8kJ $
Thus we know that the answer is $ 434.8kJ $ .

Note :
This is because all physical properties are represented by dimensions. Units are the ones which are allocated to the dimensions. There are two types of dimensions namely primary or fundamental and secondary or derived dimensions. As per the law of energy conservation, it is known that change in internal energy is equal to heat transferred to, less the work done by, the system.