Consider the following data:
$(i)$ $2Al(s) + 6HCl(aq) \to Al_2Cl_6(aq) + 3H_2(g) + 1200 \text{ kJ/mol}$
(ii) $H_2(g) + Cl_2(g) \to 2HCl(g) + 164 \text{ kJ/mol}$
(iii) $HCl(g) + aq \to HCl(aq) + 83 \text{ kJ/mol}$
(iv) $Al_2Cl_6(s) + aq \to Al_2Cl_6(aq) + 663 \text{ kJ/mol}$
The enthalpy of formation of anhydrous solid $Al_2Cl_6$ is:
$(i)$ $2Al(s) + 6HCl(aq) \to Al_2Cl_6(aq) + 3H_2(g) + 1200 \text{ kJ/mol}$
(ii) $H_2(g) + Cl_2(g) \to 2HCl(g) + 164 \text{ kJ/mol}$
(iii) $HCl(g) + aq \to HCl(aq) + 83 \text{ kJ/mol}$
(iv) $Al_2Cl_6(s) + aq \to Al_2Cl_6(aq) + 663 \text{ kJ/mol}$
The enthalpy of formation of anhydrous solid $Al_2Cl_6$ is:
- A$-648 \text{ kJ mol}^{-1}$
- B$-1350 \text{ kJ mol}^{-1}$
- C$-2002 \text{ kJ mol}^{-1}$
- D$-1527 \text{ kJ mol}^{-1}$
Explore More
Similar Questions
Calculate the enthalpy of formation of nitric oxide $(NO)$ in $kJ \, mol^{-1}$ from the following data:
$NO_{(g)} + CO_{(g)} \rightarrow \frac{1}{2} N_{2(g)} + CO_{2(g)}; \Delta H^o = -372.2 \, kJ \, mol^{-1}$
$\Delta H_f^o (CO) = -110.5 \, kJ \, mol^{-1}$
$\Delta H_f^o (CO_2) = -393.5 \, kJ \, mol^{-1}$
$NO_{(g)} + CO_{(g)} \rightarrow \frac{1}{2} N_{2(g)} + CO_{2(g)}; \Delta H^o = -372.2 \, kJ \, mol^{-1}$
$\Delta H_f^o (CO) = -110.5 \, kJ \, mol^{-1}$
$\Delta H_f^o (CO_2) = -393.5 \, kJ \, mol^{-1}$
Medium
View SolutionBased on the following thermochemical equations:
$H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)}; \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2}O_{2(g)} \to CO_{2(g)}; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}; \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta H = X \ kJ$
The value of $X$ is ...... $kJ$.
$H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)}; \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2}O_{2(g)} \to CO_{2(g)}; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}; \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta H = X \ kJ$
The value of $X$ is ...... $kJ$.
MediumAIPMT 1992
View SolutionThe enthalpy of combustion of methane,graphite and dihydrogen at $298 \, K$ are $-890.3 \, kJ \, mol^{-1}$,$-393.5 \, kJ \, mol^{-1}$ and $-285.8 \, kJ \, mol^{-1}$ respectively. The enthalpy of formation of $CH_{4(g)}$ will be:
$(i) -74.8 \, kJ \, mol^{-1}$
$(ii) -52.27 \, kJ \, mol^{-1}$
$(iii) +74.8 \, kJ \, mol^{-1}$
$(iv) +52.26 \, kJ \, mol^{-1}$
$(i) -74.8 \, kJ \, mol^{-1}$
$(ii) -52.27 \, kJ \, mol^{-1}$
$(iii) +74.8 \, kJ \, mol^{-1}$
$(iv) +52.26 \, kJ \, mol^{-1}$
Medium
View SolutionThe heat evolved in the combustion of benzene is given by $C_6H_6 + 7.5 O_2 \to 6CO_2(g) + 3H_2O(l); \Delta H = -3264.6 \, kJ$. Which of the following quantities of heat energy will be evolved when $39 \, g$ of $C_6H_6$ is burnt? (in $kJ$)
Medium
View SolutionWhat type of process is the formation of ozone from oxygen?
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