Comment on the thermodynamic stability of $NO_{(g)}$,given:
$\frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow NO_{(g)}; \Delta_r H^{\ominus} = 90 \, kJ \, mol^{-1}$
$NO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow NO_{2(g)}; \Delta_r H^{\ominus} = -74 \, kJ \, mol^{-1}$
$\frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow NO_{(g)}; \Delta_r H^{\ominus} = 90 \, kJ \, mol^{-1}$
$NO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow NO_{2(g)}; \Delta_r H^{\ominus} = -74 \, kJ \, mol^{-1}$
(N/A) The positive value of $\Delta_r H^{\ominus} = 90 \, kJ \, mol^{-1}$ indicates that heat is absorbed during the formation of $NO_{(g)}$ from its elements.
This means that $NO_{(g)}$ has higher energy than the reactants ($N_{2(g)}$ and $O_{2(g)}$),making it thermodynamically unstable.
Conversely,the negative value of $\Delta_r H^{\ominus} = -74 \, kJ \, mol^{-1}$ for the reaction $NO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow NO_{2(g)}$ indicates that heat is evolved,leading to a more stable product $NO_{2(g)}$.
Therefore,$NO_{(g)}$ is thermodynamically unstable and tends to convert into $NO_{2(g)}$.
This means that $NO_{(g)}$ has higher energy than the reactants ($N_{2(g)}$ and $O_{2(g)}$),making it thermodynamically unstable.
Conversely,the negative value of $\Delta_r H^{\ominus} = -74 \, kJ \, mol^{-1}$ for the reaction $NO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow NO_{2(g)}$ indicates that heat is evolved,leading to a more stable product $NO_{2(g)}$.
Therefore,$NO_{(g)}$ is thermodynamically unstable and tends to convert into $NO_{2(g)}$.
Explore More
Similar Questions
When the aqueous solution of $0.5 \, mole$ of $HNO_3$ is mixed with $0.3 \, mole$ of $OH^{-}$ solution,what will be the liberated heat in $kJ$? (Enthalpy of neutralization is $= 57.1 \, kJ \, mol^{-1}$)
Medium
View SolutionThe enthalpy changes for the following processes are listed below:
$Cl_{2(g)} \to 2Cl_{(g)}$,$\Delta H = 242.3 \ kJ \ mol^{-1}$
$I_{2(g)} \to 2I_{(g)}$,$\Delta H = 151.0 \ kJ \ mol^{-1}$
$ICl_{(g)} \to I_{(g)} + Cl_{(g)}$,$\Delta H = 211.3 \ kJ \ mol^{-1}$
$I_{2(s)} \to I_{2(g)}$,$\Delta H = 62.76 \ kJ \ mol^{-1}$
Given that the standard states for iodine and chlorine are $I_{2(s)}$ and $Cl_{2(g)}$,the standard enthalpy of formation for $ICl_{(g)}$ is .............. $kJ \ mol^{-1}$
$Cl_{2(g)} \to 2Cl_{(g)}$,$\Delta H = 242.3 \ kJ \ mol^{-1}$
$I_{2(g)} \to 2I_{(g)}$,$\Delta H = 151.0 \ kJ \ mol^{-1}$
$ICl_{(g)} \to I_{(g)} + Cl_{(g)}$,$\Delta H = 211.3 \ kJ \ mol^{-1}$
$I_{2(s)} \to I_{2(g)}$,$\Delta H = 62.76 \ kJ \ mol^{-1}$
Given that the standard states for iodine and chlorine are $I_{2(s)}$ and $Cl_{2(g)}$,the standard enthalpy of formation for $ICl_{(g)}$ is .............. $kJ \ mol^{-1}$
MediumAIIMS 2015
View SolutionGiven that $:$
$2 C_{(s)} + 2 O_{2_{(g)}} \rightarrow 2 CO_{2_{(g)}} ; \Delta H = -787 \ kJ$
$H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \rightarrow H_2 O_{(l)} ; \Delta H = -286 \ kJ$
$C_2 H_{2_{(g)}} + \frac{5}{2} O_{2_{(g)}} \rightarrow 2 CO_{2_{(g)}} + H_2 O_{(l)} ; \Delta H = -1301 \ kJ$
The heat of formation of acetylene will be $:-$
$2 C_{(s)} + 2 O_{2_{(g)}} \rightarrow 2 CO_{2_{(g)}} ; \Delta H = -787 \ kJ$
$H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \rightarrow H_2 O_{(l)} ; \Delta H = -286 \ kJ$
$C_2 H_{2_{(g)}} + \frac{5}{2} O_{2_{(g)}} \rightarrow 2 CO_{2_{(g)}} + H_2 O_{(l)} ; \Delta H = -1301 \ kJ$
The heat of formation of acetylene will be $:-$
Medium
View SolutionCalculate the standard enthalpy of formation of $CH_{3}OH_{(l)}$ from the following data:
$CH_{3}OH_{(l)} + \frac{3}{2} O_{2_{(g)}} \rightarrow CO_{2_{(g)}} + 2 H_{2}O_{(l)}$; $\Delta_{r} H^{\ominus} = -726 \ kJ \ mol^{-1}$
$C_{(graphite)} + O_{2_{(g)}} \rightarrow CO_{2_{(g)}}$; $\Delta_{c} H^{\ominus} = -393 \ kJ \ mol^{-1}$
$H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \rightarrow H_{2}O_{(l)}$; $\Delta_{f} H^{\ominus} = -286 \ kJ \ mol^{-1}$
$CH_{3}OH_{(l)} + \frac{3}{2} O_{2_{(g)}} \rightarrow CO_{2_{(g)}} + 2 H_{2}O_{(l)}$; $\Delta_{r} H^{\ominus} = -726 \ kJ \ mol^{-1}$
$C_{(graphite)} + O_{2_{(g)}} \rightarrow CO_{2_{(g)}}$; $\Delta_{c} H^{\ominus} = -393 \ kJ \ mol^{-1}$
$H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \rightarrow H_{2}O_{(l)}$; $\Delta_{f} H^{\ominus} = -286 \ kJ \ mol^{-1}$
Medium
View SolutionThe heat of neutralization of which acid and base is close to $-13.6 \ kcal/mol$?
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo