For the real gases reaction $ 2C{O_{\left( g \right)}} + {O_{2\left( g \right)}} \to 2C{O_{2\left( g \right)}} $ : $ \Delta {\rm H} = - 560 $ $ kJ $ .In $ 10 $ litre rigid vessel at $ 500K $ .The initial pressure is $ 70 $ bar and after the reaction it becomes $ 40 $ bar. The change in internal energy is
(A) $ - 557 $ $ kJ $
(B) $ - 530 $ $ kJ $
(C) $ - 563 $ $ kJ $
(D) None of these
Answer
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Hint :Internal energy is composed of kinetic energy as well as potential energy of the constituents present in the system. It is represented as $ \left( U \right) $ .Value of internal energy is either positive or negative depending upon the reaction condition.
Complete Step By Step Answer:
Enthalpy is related to internal energy of the system, pressure of the system and volume of the system. The mathematical expression for such relationship is
$ {\rm H} = U + pV $
Where $ H $ is enthalpy of the system
$ U $ Is internal energy of system
$ p $ Pressure of system
$ V $ Volume of system.
From the given problem we see that reaction is carried out in a vessel of fixed capacity of $ 10 $ litre. Hence the reaction proceeds under the isochoric condition.
Therefore, the equation modified and become-
$ \Delta H = \Delta U + \left( {{p_1} - {p_2}} \right)V $
Where $ {p_1} $ is initial pressure of system as given $ 70 $ bar
$ {p_2} $ is final pressure of system as given $ 40 $ bar
$ \Delta {\rm H} = - 560 $ $ kJ $ is enthalpy of system
$ V $ is volume of system as given $ 10 $ litre
Now put all the values in the equation
$ - 560 = \Delta U + \left( {70 - 40} \right)10 $
$ - 560 = \Delta U + \left( {30} \right)10 $
On simplifying the equation, we get
$ - 560 = \Delta U + 300 $
Convert the value of $ 300 $ litre-atmosphere into $ 300kJ $ by multiplying $ \left( {0.1} \right) $
$ - 560 = \Delta U + 300 \times 0.1 $
Rearrange the equation in terms of internal energy of system
$ \Delta U = - 560 + 30 $
$ \Delta U = 530 $ $ kJ $
Hence, the change in internal energy is $ 530kJ $ .Therefore, option $ \left( 2 \right) $ is the correct option.
Note :
Internal energy is a type of state function which is dependent upon the initial and final states of the chemical reaction.
Reaction process under constant volume condition is known as isochoric process.
Complete Step By Step Answer:
Enthalpy is related to internal energy of the system, pressure of the system and volume of the system. The mathematical expression for such relationship is
$ {\rm H} = U + pV $
Where $ H $ is enthalpy of the system
$ U $ Is internal energy of system
$ p $ Pressure of system
$ V $ Volume of system.
From the given problem we see that reaction is carried out in a vessel of fixed capacity of $ 10 $ litre. Hence the reaction proceeds under the isochoric condition.
Therefore, the equation modified and become-
$ \Delta H = \Delta U + \left( {{p_1} - {p_2}} \right)V $
Where $ {p_1} $ is initial pressure of system as given $ 70 $ bar
$ {p_2} $ is final pressure of system as given $ 40 $ bar
$ \Delta {\rm H} = - 560 $ $ kJ $ is enthalpy of system
$ V $ is volume of system as given $ 10 $ litre
Now put all the values in the equation
$ - 560 = \Delta U + \left( {70 - 40} \right)10 $
$ - 560 = \Delta U + \left( {30} \right)10 $
On simplifying the equation, we get
$ - 560 = \Delta U + 300 $
Convert the value of $ 300 $ litre-atmosphere into $ 300kJ $ by multiplying $ \left( {0.1} \right) $
$ - 560 = \Delta U + 300 \times 0.1 $
Rearrange the equation in terms of internal energy of system
$ \Delta U = - 560 + 30 $
$ \Delta U = 530 $ $ kJ $
Hence, the change in internal energy is $ 530kJ $ .Therefore, option $ \left( 2 \right) $ is the correct option.
Note :
Internal energy is a type of state function which is dependent upon the initial and final states of the chemical reaction.
Reaction process under constant volume condition is known as isochoric process.