NEET Chemistry MCQs

Increase in entropy due to increase in gaseous moles.

Decrease in entropy due to decrease in gaseous moles.

Constant entropy as gaseous moles remain the same.

The reaction is exothermic.

Adsorption is generally an exothermic process

ΔG (Gibbs free energy change) is negative

ΔS (entropy change) is negative

ΔS (entropy change) is positive

Entropy increases

Enthalpy increases

Free energy increases

Entropy decreases

The entropy change for the reaction is negative.

The reaction becomes less spontaneous at higher temperatures.

The free energy change becomes more negative at higher temperatures.

The enthalpy change for the reaction is positive.

A larger increase in entropy during the reaction.

A smaller increase in entropy during the reaction.

A more negative enthalpy change for the reaction.

A less negative enthalpy change for the reaction.

The number of gaseous molecules increases during the reaction.

The reaction is exothermic.

Carbon is a good reducing agent.

Carbon monoxide is a stable gas.

Growth and development

Response to stimuli

Entropy always decreases

Adaptation through evolution

$\Delta S = {\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} imes nF$

${\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} = \frac{{\Delta G - \Delta H}}{T}$

${\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} = \frac{{\partial \left( {\Delta S} \right)}}{{\partial T}}$

$- \Delta S = {\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} imes nF$

$\Delta S = {\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} imes nF$

${\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} = \frac{{\Delta G - \Delta H}}{T}$

${\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} = \frac{{\partial \left( {\Delta S} \right)}}{{\partial T}}$

$- \Delta S = {\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} imes nF$

Enthalpy of mixing is zero

Entropy of mixing is zero

Free energy of mixing is zero

Free energy as well as the entropy of mixing are each zero