The 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 .............. $kJ \ mol^{-1}$
- A$+16.8$
- B$+244.8$
- C$-14.6$
- D$-16.8$
Explore More
Similar Questions
Energy required to dissociate $4 \ g$ of gaseous hydrogen into free gaseous atoms is $208 \ kcal$ at $25 \ ^oC$. The bond energy of $H-H$ bond will be
Medium
View SolutionCalculate the standard enthalpy change of the following reaction: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(\ell)}$ if $\Delta_{f} H^{\circ}(CH_{4}) = -75 \ kJ \ mol^{-1}$,$\Delta_{f} H^{\circ}(CO_{2}) = -390 \ kJ \ mol^{-1}$,and $\Delta_{f} H^{\circ}(H_{2}O) = -286 \ kJ \ mol^{-1}$.
Given $:$
$\Delta H^{\ominus}_{sub}[C(graphite)] = 710 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{C-H} = 414 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{H-H} = 436 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{C=C} = 611 \ kJ \ mol^{-1}$
The $\Delta H^{\ominus}_{f}$ for $CH_2=CH_2$ is $............ \ kJ \ mol^{-1}$ $(nearest \ integer \ value)$
$\Delta H^{\ominus}_{sub}[C(graphite)] = 710 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{C-H} = 414 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{H-H} = 436 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{C=C} = 611 \ kJ \ mol^{-1}$
The $\Delta H^{\ominus}_{f}$ for $CH_2=CH_2$ is $............ \ kJ \ mol^{-1}$ $(nearest \ integer \ value)$
Combustion of $1$ $mol$ of benzene is expressed as:
$C_6H_{6(l)} + \frac{15}{2} O_{2(g)} \rightarrow 6CO_{2(g)} + 3H_2O_{(l)}$
The standard enthalpy of combustion of $2$ $mol$ of benzene is $-x$ $kJ$.
$x = . . . . . . . . . .$
$(1)$ Standard enthalpy of formation of $1$ $mol$ of $C_6H_{6(l)}$ is $48.5$ $kJ \ mol^{-1}$.
$(2)$ Standard enthalpy of formation of $1$ $mol$ of $CO_{2(g)}$ is $-393.5$ $kJ \ mol^{-1}$.
$(3)$ Standard enthalpy of formation of $1$ $mol$ of $H_2O_{(l)}$ is $-286$ $kJ \ mol^{-1}$.
$C_6H_{6(l)} + \frac{15}{2} O_{2(g)} \rightarrow 6CO_{2(g)} + 3H_2O_{(l)}$
The standard enthalpy of combustion of $2$ $mol$ of benzene is $-x$ $kJ$.
$x = . . . . . . . . . .$
$(1)$ Standard enthalpy of formation of $1$ $mol$ of $C_6H_{6(l)}$ is $48.5$ $kJ \ mol^{-1}$.
$(2)$ Standard enthalpy of formation of $1$ $mol$ of $CO_{2(g)}$ is $-393.5$ $kJ \ mol^{-1}$.
$(3)$ Standard enthalpy of formation of $1$ $mol$ of $H_2O_{(l)}$ is $-286$ $kJ \ mol^{-1}$.
The enthalpy changes at $298 \ K$ in successive breaking of $O-H$ bonds of $H_2O$ are:
$H_2O_{(g)} \to H_{(g)} + OH_{(g)}, \Delta H = 498 \ kJ \ mol^{-1}$
$OH_{(g)} \to H_{(g)} + O_{(g)}, \Delta H = 428 \ kJ \ mol^{-1}$
The bond enthalpy of the $O-H$ bond is ..... $kJ \ mol^{-1}$.
$H_2O_{(g)} \to H_{(g)} + OH_{(g)}, \Delta H = 498 \ kJ \ mol^{-1}$
$OH_{(g)} \to H_{(g)} + O_{(g)}, \Delta H = 428 \ kJ \ mol^{-1}$
The bond enthalpy of the $O-H$ bond is ..... $kJ \ mol^{-1}$.
Difficult
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