For the reaction $H_{2} + I_{2} \rightarrow 2 HI$,provide the potential energy profile diagram and prove that it is an exothermic reaction.
(N/A) The potential energy profile diagram shows the energy change during the reaction.
From the graph,it is clear that the potential energy of the products $(E_{P})$ is less than the potential energy of the reactants $(E_{R})$.
Since $H_{P} < H_{R}$,the change in enthalpy is given by:
$\Delta H = (H_{P} - H_{R}) < 0$ (negative value).
Because energy is released during the process,the reaction is exothermic.
From the graph,it is clear that the potential energy of the products $(E_{P})$ is less than the potential energy of the reactants $(E_{R})$.
Since $H_{P} < H_{R}$,the change in enthalpy is given by:
$\Delta H = (H_{P} - H_{R}) < 0$ (negative value).
Because energy is released during the process,the reaction is exothermic.
Explore More
Similar Questions
The energy required to break one mole of $Cl-Cl$ bonds in $Cl_2$ is $242 \, kJ \, mol^{-1}.$ The longest wavelength of light capable of breaking a single $Cl-Cl$ bond is $............ \, nm.$
$(c = 3 \times 10^8 \, m \, s^{-1}$ and $N_A = 6.02 \times 10^{23} \, mol^{-1}).$
$(c = 3 \times 10^8 \, m \, s^{-1}$ and $N_A = 6.02 \times 10^{23} \, mol^{-1}).$
MediumAIEEE 2010
View SolutionThe enthalpies of combustion of $C$(graphite) and $C$(diamond) are $-393.8 \ kJ \ mol^{-1}$ and $-395.3 \ kJ \ mol^{-1}$ respectively. The enthalpy of conversion of $C$(graphite) to $C$(diamond) is
When equal volumes of molar hydrochloric acid $(HCl)$ and sulfuric acid $(H_2SO_4)$ are neutralized by a dilute $NaOH$ solution,the heat produced is $x \ KCal$ and $y \ KCal$ respectively. Which of the following is correct?
Medium
View SolutionWhat is the standard enthalpy of reaction (in $kJ$) when two moles of $Fe_2O_{3(s)}$ react with $H_2$ gas to give $Fe$ metal? $\Delta H_f^{\circ}$ of $Fe_2O_{3(s)}$ and $H_2O_{(l)}$ are $-824.2$ and $-285.83 \ kJ \ mol^{-1}$ respectively.
Calculate the standard enthalpy change for the synthesis of ammonia gas from the following data:
$i$. $2 H_{2(g)} + N_{2(g)} \longrightarrow N_{2}H_{4(g)}$; $\Delta_{r}H_{1}^{0} = 95.4 \ kJ$
$ii$. $N_{2}H_{4(g)} + H_{2(g)} \longrightarrow 2 NH_{3(g)}$; $\Delta_{r}H_{2}^{0} = -187.6 \ kJ$ (in $kJ$)
$i$. $2 H_{2(g)} + N_{2(g)} \longrightarrow N_{2}H_{4(g)}$; $\Delta_{r}H_{1}^{0} = 95.4 \ kJ$
$ii$. $N_{2}H_{4(g)} + H_{2(g)} \longrightarrow 2 NH_{3(g)}$; $\Delta_{r}H_{2}^{0} = -187.6 \ kJ$ (in $kJ$)
Vedclass 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