$PbO$ exists in two crystalline forms: yellow and red. The standard enthalpies of formation for these two forms are $-217.3 \ kJ/mol$ and $-219.0 \ kJ/mol$,respectively. Calculate the enthalpy of transition for the process: $PbO \text{ (yellow)} \rightarrow PbO \text{ (red)}$ in $kJ/mol$.

  • A
    $-2.3$
  • B
    $-1.3$
  • C
    $-1.7$
  • D
    $-2.5$

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Similar Questions

Calculate the enthalpy of formation of sucrose $(C_{12}H_{22}O_{11})$ in $kJ \, mol^{-1}$ using the following data:
$(i) \, C_{12}H_{22}O_{11} + 12O_2 \to 12CO_2 + 11H_2O, \, \Delta H = -5200.7 \, kJ \, mol^{-1}$
$(ii) \, C + O_2 \to CO_2, \, \Delta H = -394.5 \, kJ \, mol^{-1}$
$(iii) \, H_2 + \frac{1}{2}O_2 \to H_2O, \, \Delta H = -285.8 \, kJ \, mol^{-1}$

For the reaction $C_{2}H_{6} \rightarrow C_{2}H_{4} + H_{2}$,the reaction enthalpy $\Delta_{r}H = \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \, kJ \, mol^{-1}$. (Round off to the Nearest Integer). [Given: Bond enthalpies in $kJ \, mol^{-1} : C-C : 347, C=C : 611, C-H : 414, H-H : 436$]

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 .............. $kJ \ mol^{-1}$

Which is the best definition of “heat of neutralization”?

Comment on the thermodynamic stability of $NO_{(g)}$,given:
$\frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow NO_{(g)}; \Delta_r H^{\ominus} = 90 \, kJ \, mol^{-1}$
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