2 , mol of zinc is dissolved in HCl at 25 , ^ | vedclass

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Question

(B) The reaction is: $Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)$.For $1 \, mol$ of $Zn$,the change in gaseous moles is $\Delta n_g = 1 \, mol \

Vedclass · Accepted Answer

$-4.955 \, kJ$

Vedclass · Answer

(B) The reaction is: $Zn(s) + 2HCl(aq) ightarrow ZnCl_2(aq) + H_2(g)$.For $1 \, mol$ of $Zn$,the change in gaseous moles is $\Delta n_g = 1 \, mol \, (H_2) - 0 = 1$.For $2 \, mol$ of $Zn$,$\Delta n_g = 2 \, mol$.The work done in an open vessel is given by $W = -\Delta n_g RT$.Given $T = 25 \, ^oC = 298 \, K$ and $R = 8.314 \, J \, K^{-1} \, mol^{-1}$.$W = -2 	imes 8.314 	imes 298 = -4955.144 \, J$.Converting to $kJ$,$W = -4.955 \, kJ$.

	$\Delta_r H^{\circ}$	$\Delta_r S^{\circ}$	$\Delta_r G^{\circ}$	Spontaneity of the reaction
$A$	$+$	$+$	$+$	Spontaneous at low $T$
$B$	$+$	$+$	$-$	Spontaneous at high $T$
$C$	$-$	$-$	$-$	Spontaneous at low $T$
$D$	$+$	$-$	$+$	Spontaneous at high $T$

$2 \, mol$ of zinc is dissolved in $HCl$ at $25 \, ^oC$. What is the work done in an open vessel?

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