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Grade upto college level Physical Chemistry

The standard heats of formation at 298 K for CCI4(g), H2O(g), CO2(g) and HCI(g) are -25.5, -57.8, -94.1 and -22.1 kcal/mol respectively. Calculate ∆ Ho298 for the reaction
CCI4(g) + 2H2O → CO2(g) + 4HCI(g)

Profile image of Shane Macguire
12 Years agoGrade upto college level
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1 Answer

Profile image of Deepak Patra
12 Years ago
Hello Student,
Please find the answer to your question
Since we know that
Heat content of the compound
= Heat of formation ∆H, \Delta H_{298}^{o}
= Total heat contents of the products
- Total heat contents of the reactants
Writing the given chemical reaction,
CCI4(g) + 2H2O(g) → CO2(g) + 4HCI(g); \Delta H_{298}^{o}= ?
\Delta H_{298}^{o}= [\Delta H_{CO_{2}}+ 4 *∆HHCI] – [\Delta H_{CCI_{4}}+ 2 *\Delta H_{H_{2}O}
Given, \Delta H_{CCI_{4}}= - 25.5 kcal/mole \Delta H_{H_{2}O}
= - 57.8 kcal/mole
\Delta H_{CO_{2}}= - 94.1 kcal/mole ∆HHCI = - 22.1 kcal/mole
Substituting the values in the above equation,
\Delta H_{298}^{o}= [-94.1 + 4 * - 22.1] – [-25.5 + 2 * - 57.8]
= [-94.1 – 88.4] – [-25.5 – 115.6]
= - 182.5 + 141.1 = -41.4 kcal

Thanks
jitender
askIITians Faculty

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