Consider the reaction at 300 K: H_{2(g)} + C | vedclass

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(C) The given reaction is $H_{2(g)} + Cl_{2(g)} 	o 2HCl_{(g)}$.First,calculate the change in the number of gaseous moles,$\Delta n_g = \sum n_{g(prod

Vedclass · Accepted Answer

$-370$

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(C) The given reaction is $H_{2(g)} + Cl_{2(g)} 	o 2HCl_{(g)}$.First,calculate the change in the number of gaseous moles,$\Delta n_g = \sum n_{g(products)} - \sum n_{g(reactants)} = 2 - (1 + 1) = 0$.Using the relation $\Delta H^o = \Delta U^o + \Delta n_g RT$,since $\Delta n_g = 0$,we have $\Delta H^o = \Delta U^o$.For the reaction of $1 \ mole$ of $H_2$ with $1 \ mole$ of $Cl_2$,$\Delta H^o = -185 \ kJ$.Since the question asks for the reaction of $2 \ moles$ of $H_2$ with $2 \ moles$ of $Cl_2$,the enthalpy change will be $2 	imes (-185 \ kJ) = -370 \ kJ$.Therefore,$\Delta U^o = -370 \ kJ$.

Consider the reaction at $300 \ K$: $H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}$; $\Delta H^o = -185 \ kJ$. If $2 \ mole$ of $H_2$ completely react with $2 \ mole$ of $Cl_2$ to form $HCl$,what is $\Delta U^o$ for this reaction in $kJ$?

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