What happens in a process if pressure and temperature keep constant?
(A) $\Delta H=\Delta E$
(B) $\Delta H=T\Delta S$
(C) $\Delta H=0$
(D) $\Delta S=0$
Answer
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Hint:re, the pressure and temperature is constant throughout the reaction, so we can say that $\Delta T$ and $\Delta P$ for the reaction will be zero. Now, we can put these in the formula of given quantities and find its value.
Complete step-by-step answer:Here, we are given that the change in pressure and temperature during the course of the reaction is zero. Let’s see the relation of temperature of pressure and temperature with given quantities in order to get the answer.
$\Delta H$ is enthalpy change and $\Delta E$ is total energy of the system. The relation between them is shown below.
$\Delta H=\Delta E+P\Delta V$
So, we can say that if pressure and temperature is constant, there will still be some change in volume of the reaction and hence in a given case $\Delta H$ cannot be equal to $\Delta E$. Thus, option (A) is not correct.
- $\Delta H$ can also be expressed by the following formula.
$\Delta H={{C}_{p}}\Delta T$ ……(1)
Where ${{C}_{p}}$ is the heat capacity of the system at constant pressure. But we are also given that reaction takes place at constant temperature. So, we can say that $\Delta T$ will be zero for the reaction. Hence, we can say that $\Delta H=0$ for that reaction from equation (1) as we put $\Delta T$ equal to zero.
- Now, we know the relation between entropy and enthalpy which can be given by
$\Delta G=\Delta H-T\Delta S$ …….(2)
So, we can say from the equation (2) that even at constant temperature and pressure, $T\Delta S>H$. Thus, option (B) is also not correct.
- We can write the formula of entropy as $\Delta S=\dfrac{\Delta Q}{T}$
So, from this, we can say that even if the temperature remains constant over the time of the reaction, $\Delta S$ cannot be zero.
So, from the above reaction, we can say that the correct option is (C) $\Delta H=0$ .
Note:Do not get confused between the internal energy ($\Delta U$) and total energy of the system ($\Delta E$). Total energy of the system involves internal energy of the system along with the kinetic and potential energy of the system.
Complete step-by-step answer:Here, we are given that the change in pressure and temperature during the course of the reaction is zero. Let’s see the relation of temperature of pressure and temperature with given quantities in order to get the answer.
$\Delta H$ is enthalpy change and $\Delta E$ is total energy of the system. The relation between them is shown below.
$\Delta H=\Delta E+P\Delta V$
So, we can say that if pressure and temperature is constant, there will still be some change in volume of the reaction and hence in a given case $\Delta H$ cannot be equal to $\Delta E$. Thus, option (A) is not correct.
- $\Delta H$ can also be expressed by the following formula.
$\Delta H={{C}_{p}}\Delta T$ ……(1)
Where ${{C}_{p}}$ is the heat capacity of the system at constant pressure. But we are also given that reaction takes place at constant temperature. So, we can say that $\Delta T$ will be zero for the reaction. Hence, we can say that $\Delta H=0$ for that reaction from equation (1) as we put $\Delta T$ equal to zero.
- Now, we know the relation between entropy and enthalpy which can be given by
$\Delta G=\Delta H-T\Delta S$ …….(2)
So, we can say from the equation (2) that even at constant temperature and pressure, $T\Delta S>H$. Thus, option (B) is also not correct.
- We can write the formula of entropy as $\Delta S=\dfrac{\Delta Q}{T}$
So, from this, we can say that even if the temperature remains constant over the time of the reaction, $\Delta S$ cannot be zero.
So, from the above reaction, we can say that the correct option is (C) $\Delta H=0$ .
Note:Do not get confused between the internal energy ($\Delta U$) and total energy of the system ($\Delta E$). Total energy of the system involves internal energy of the system along with the kinetic and potential energy of the system.
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