(B) The change in entropy for the reaction is given by: $\Delta S = \sum \Delta S_{\text{products}} - \sum \Delta S_{\text{reactants}}$$\Delta S = \De

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

(B) The change in entropy for the reaction is given by: $\Delta S = \sum \Delta S_{\text{products}} - \sum \Delta S_{\text{reactants}}$$\Delta S = \Delta S_{XY_3} - [\frac{1}{2} \Delta S_{X_2} + \frac{3}{2} \Delta S_{Y_2}]$$\Delta S = 50 - [\frac{60}{2} + \frac{3 \times 40}{2}] \ J/mol \cdot K$$\Delta S = 50 - [30 + 60] = 50 - 90 = -40 \ J/mol \cdot K = -0.04 \ kJ/mol \cdot K$At equilibrium,$\Delta G = 0$,so $\Delta G = \Delta H - T \Delta S = 0$,which implies $T = \frac{\Delta H}{\Delta S}$.$T = \frac{-30 \ kJ/mol}{-0.04 \ kJ/mol \cdot K} = 750 \ K$.

Class 11 Chemistry Thermodynamics Mix Examples-Thermodynamics and Thermochemistry

Difficult

For the reaction $\frac{1}{2}X_2 + \frac{3}{2}Y_2 \to XY_3$,$\Delta H = -30 \ kJ/mol$. Given $\Delta S_{X_2} = 60 \ J/mol \cdot K$,$\Delta S_{Y_2} = 40 \ J/mol \cdot K$,and $\Delta S_{XY_3} = 50 \ J/mol \cdot K$,calculate the temperature at equilibrium in $K$.

A
$500$
B
$750$
C
$1000$
D
$1250$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Chemistry Questions

Chapter

Thermodynamics

Subtopic

Mix Examples-Thermodynamics and Thermochemistry

Match the following processes with their corresponding entropy changes:

Process Entropy Change

$(a)$ Liquid to vapor conversion $(1)$ $\Delta S = 0$

$(b)$ Process not spontaneous at any temperature $(2)$ $\Delta S = (+)$

$(c)$ Reversible expansion of an ideal gas $(3)$ $\Delta S = (-)$

Medium

View Solution

The molar heat capacity for an ideal gas at constant pressure is $20.785 \ J \ K^{-1} \ mol^{-1}$. The change in internal energy is $5000 \ J$ upon heating it from $300 \ K$ to $500 \ K$. The number of moles of the gas is [Nearest integer] (Given: $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

MediumJEE MAIN 2022

View Solution

In a constant pressure process for $5 \text{ mole}$ of an ideal monoatomic gas,the temperature of the gas increased from $300 \ K$ to $500 \ K$. Which of the following is incorrect regarding the process?

Difficult

View Solution

Which of the following relations between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is correct for the given reactions?

Medium

View Solution

Match the following items in List-$I$ with their corresponding expressions in List-$II$.
List-$I$ List-$II$
$A$. At constant volume the change in internal energy of a system $I$. $W = -2.303 nRT \log \frac{V_f}{V_i}$
$B$. Isothermal irreversible change $II$. $W_{adiabatic} = \Delta U$
$C$. Isothermal reversible change $III$. $q_V = \Delta U$
$D$. Adiabatic change $IV$. $W = -p_{ex} (V_f - V_i)$
$V$. $\Delta U = \Delta H - \Delta nRT$

MediumAP EAMCET 2018

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Process	Entropy Change
$(a)$ Liquid to vapor conversion	$(1)$ $\Delta S = 0$
$(b)$ Process not spontaneous at any temperature	$(2)$ $\Delta S = (+)$
$(c)$ Reversible expansion of an ideal gas	$(3)$ $\Delta S = (-)$

List-$I$	List-$II$
$A$. At constant volume the change in internal energy of a system	$I$. $W = -2.303 nRT \log \frac{V_f}{V_i}$
$B$. Isothermal irreversible change	$II$. $W_{adiabatic} = \Delta U$
$C$. Isothermal reversible change	$III$. $q_V = \Delta U$
$D$. Adiabatic change	$IV$. $W = -p_{ex} (V_f - V_i)$
	$V$. $\Delta U = \Delta H - \Delta nRT$

For the reaction $\frac{1}{2}X_2 + \frac{3}{2}Y_2 \to XY_3$,$\Delta H = -30 \ kJ/mol$. Given $\Delta S_{X_2} = 60 \ J/mol \cdot K$,$\Delta S_{Y_2} = 40 \ J/mol \cdot K$,and $\Delta S_{XY_3} = 50 \ J/mol \cdot K$,calculate the temperature at equilibrium in $K$.

Similar Questions

Match the following processes with their corresponding entropy changes: Process Entropy Change $(a)$ Liquid to vapor conversion $(1)$ $\Delta S = 0$ $(b)$ Process not spontaneous at any temperature $(2)$ $\Delta S = (+)$ $(c)$ Reversible expansion of an ideal gas $(3)$ $\Delta S = (-)$

The molar heat capacity for an ideal gas at constant pressure is $20.785 \ J \ K^{-1} \ mol^{-1}$. The change in internal energy is $5000 \ J$ upon heating it from $300 \ K$ to $500 \ K$. The number of moles of the gas is [Nearest integer] (Given: $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

In a constant pressure process for $5 \text{ mole}$ of an ideal monoatomic gas,the temperature of the gas increased from $300 \ K$ to $500 \ K$. Which of the following is incorrect regarding the process?

Which of the following relations between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is correct for the given reactions?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Match the following processes with their corresponding entropy changes:

Process Entropy Change

$(a)$ Liquid to vapor conversion $(1)$ $\Delta S = 0$

$(b)$ Process not spontaneous at any temperature $(2)$ $\Delta S = (+)$

$(c)$ Reversible expansion of an ideal gas $(3)$ $\Delta S = (-)$