The process with negative entropy change is

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Entropy change $(\Delta S)$ is related to the change in the number of moles of gaseous species. $A$ decrease in the number of gaseous moles leads

Vedclass · Accepted Answer

Synthesis of ammonia from $N_2$ and $H_2$

Vedclass · Answer

(D) Entropy change $(\Delta S)$ is related to the change in the number of moles of gaseous species. $A$ decrease in the number of gaseous moles leads to a negative entropy change.$A$: $CaSO_{4(s)} 	o CaO_{(s)} + SO_{3(g)}$ (Number of gaseous moles increases from $0$ to $1$,$\Delta S > 0$)$B$: $CO_{2(s)} 	o CO_{2(g)}$ (Solid to gas transition,$\Delta S > 0$)$C$: $I_{2(s)} 	o I_{2(aq)}$ (Solid to aqueous transition,$\Delta S > 0$)$D$: $N_{2(g)} + 3H_{2(g)} 	o 2NH_{3(g)}$ (Number of gaseous moles decreases from $4$ to $2$,$\Delta S Therefore,the synthesis of ammonia results in a decrease in entropy.

Similar Questions

The enthalpy of vaporisation of a liquid is $30 \ kJ \ mol^{-1}$ and entropy of vaporisation is $75 \ J \ K^{-1} \ mol^{-1}$. Calculate the boiling point of the liquid at $1 \ atm$. (in $K$)

The entropy values (in $J \, K^{-1} \, mol^{-1}$) of $H_{2(g)} = 130.6$,$Cl_{2(g)} = 223.0$,and $HCl_{(g)} = 186.7$ at $298 \, K$ and $1 \, atm$ pressure. Then entropy change for the reaction $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}$ is

For which of the following reactions will $\Delta S$ be maximum?

For the conversion $C$ (graphite) $\to C$ (diamond),the $\Delta S$ is:

For a spontaneous process,the correct statement is

Vedclass Test Series

Exam Paper Generator

Online Exam Module