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Question: For the real gases reaction \( 2C{O_{\left( g \right)}} + {O_{2\left( g \right)}} \to 2C{O_{2\left( ...

For the real gases reaction 2CO(g)+O2(g)2CO2(g)2C{O_{\left( g \right)}} + {O_{2\left( g \right)}} \to 2C{O_{2\left( g \right)}} : ΔH=560\Delta {\rm H} = - 560 kJkJ .In 1010 litre rigid vessel at 500K500K .The initial pressure is 7070 bar and after the reaction it becomes 4040 bar. The change in internal energy is
(A) 557- 557 kJkJ
(B) 530- 530 kJkJ
(C) 563- 563 kJkJ
(D) None of these

Explanation

Solution

Hint : Internal energy is composed of kinetic energy as well as potential energy of the constituents present in the system. It is represented as (U)\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
H=U+pV{\rm H} = U + pV
Where HH is enthalpy of the system
UU Is internal energy of system
pp Pressure of system
VV Volume of system.
From the given problem we see that reaction is carried out in a vessel of fixed capacity of 1010 litre. Hence the reaction proceeds under the isochoric condition.
Therefore, the equation modified and become-
ΔH=ΔU+(p1p2)V\Delta H = \Delta U + \left( {{p_1} - {p_2}} \right)V
Where p1{p_1} is initial pressure of system as given 7070 bar
p2{p_2} is final pressure of system as given 4040 bar
ΔH=560\Delta {\rm H} = - 560 kJkJ is enthalpy of system
VV is volume of system as given 1010 litre
Now put all the values in the equation
560=ΔU+(7040)10- 560 = \Delta U + \left( {70 - 40} \right)10
560=ΔU+(30)10- 560 = \Delta U + \left( {30} \right)10
On simplifying the equation, we get
560=ΔU+300- 560 = \Delta U + 300
Convert the value of 300300 litre-atmosphere into 300kJ300kJ by multiplying (0.1)\left( {0.1} \right)
560=ΔU+300×0.1- 560 = \Delta U + 300 \times 0.1
Rearrange the equation in terms of internal energy of system
ΔU=560+30\Delta U = - 560 + 30
ΔU=530\Delta U = 530 kJkJ
Hence, the change in internal energy is 530kJ530kJ .Therefore, option (2)\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.