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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

Calculate the standard enthalpy change for a reaction: \[\] $CO_2\: (g) + H_2\: (g) \rightarrow CO\: (g) + H_2O\: (g)$ given that $ \Delta H_f^o\: for\: CO_2\: (g), CO\: (g)\: and \: H_2O\: (g)\: as\: -393.5, -110.5\: and\: -241.8 \: kJ/mol$ respectively.

$\begin {array} {1 1} (a)\; -41.2\: kJ & \quad (b)\;745.8 \: kJ \\ (c)\;41.2 \: kJ & \quad (d)\;-745.8\: kJ \end {array}$

 

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1 Answer

$ \Delta H^o = \sum \Delta H_f^o$ (products) $ - \sum \Delta H_f^o$(reactants)
$= [ \Delta H_f^o(H_2O) + \Delta H_f^o(CO)] – [\Delta H_f^o(CO_2) + \Delta H_f^o(H_2)]$
Substituting the given values,
$ \Delta H^o = [-241.8 – 110.5] – [-393.5 + 0] = -352.3 + 393.5 = 41.2 \: kJ$
Ans : (c)
answered Mar 13, 2014 by thanvigandhi_1
 

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