For ${K_p}$ of a reaction at $300K$ is $6atm$ and $2atm$ at $450K$ . Which of the following statements is incorrect about this reaction, if $\Delta {n_g} = 1?$
A.The reaction is exothermic
B.The rate of backward reaction increases more than that of forward reaction with increase in temperature
C.${E_a}$ for the forward reaction is more than that of backward reaction
D.The difference between heat of reaction at constant pressure and that at constant volume is $RT$
Answer
608.1k+ views
Hint: We have to remember that the product of its pressure and the volume and the sum of the system’s internal energy is defined as enthalpy. It is a thermodynamic system of property. Enthalpy is represented in $H$ .
$H = U + pV$
Where,
$U$ is internal energy, $p$ is pressure and $V$ is the volume in above expression.
Complete answer:
We must remember that in thermodynamics a process or a reaction, which evolves heat in the form of heat to its surroundings from the system is known as exothermic reaction.
We know that, $\Delta H = \Delta E + P\Delta V$.
The change in enthalpy $(\Delta H)$ , is defined as the sum of the internal of the system increases and the pressure-volume work done i.e., the work is expansion.
$P\Delta V$ replaced by $\Delta {n_g}RT$ and the equation is,
$\Delta H = \Delta E + \Delta {n_g}RT$
Since $\Delta {n_g} = 1$ is given in the question, so we substitute this value we get, \[\Delta H - \Delta E = RT\] which is equal to constant. For this type of reaction the activation energy $({E_a})$ of forward reaction is lesser than the backward reaction. i.e.,${({E_a})_f} < {({E_a})_b}$
Where, f and b are forward and backward reactions respectively.
So, option C. ${E_a}$ for the forward reaction is more than that of backward reaction is the correct answer.
But, in an exothermic reaction the value of equilibrium constant ${K_p}$ decreases when the temperature is increased.
Therefore, the option C is the correct answer.
Note:
We have to remember that the changes in enthalpy $(\Delta H)$ , which can be easily measured experimentally in the thermodynamic process. If the reaction is exothermic then the change in enthalpy for the reaction is negative and if the reaction is endothermic (heat is absorbed) then the change in enthalpy is positive.
$H = U + pV$
Where,
$U$ is internal energy, $p$ is pressure and $V$ is the volume in above expression.
Complete answer:
We must remember that in thermodynamics a process or a reaction, which evolves heat in the form of heat to its surroundings from the system is known as exothermic reaction.
We know that, $\Delta H = \Delta E + P\Delta V$.
The change in enthalpy $(\Delta H)$ , is defined as the sum of the internal of the system increases and the pressure-volume work done i.e., the work is expansion.
$P\Delta V$ replaced by $\Delta {n_g}RT$ and the equation is,
$\Delta H = \Delta E + \Delta {n_g}RT$
Since $\Delta {n_g} = 1$ is given in the question, so we substitute this value we get, \[\Delta H - \Delta E = RT\] which is equal to constant. For this type of reaction the activation energy $({E_a})$ of forward reaction is lesser than the backward reaction. i.e.,${({E_a})_f} < {({E_a})_b}$
Where, f and b are forward and backward reactions respectively.
So, option C. ${E_a}$ for the forward reaction is more than that of backward reaction is the correct answer.
But, in an exothermic reaction the value of equilibrium constant ${K_p}$ decreases when the temperature is increased.
Therefore, the option C is the correct answer.
Note:
We have to remember that the changes in enthalpy $(\Delta H)$ , which can be easily measured experimentally in the thermodynamic process. If the reaction is exothermic then the change in enthalpy for the reaction is negative and if the reaction is endothermic (heat is absorbed) then the change in enthalpy is positive.
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