Question

If $\Delta H_f^0$ for $A{g}^{+}$($\infty$ diluted), $N{O}_3^{-}$ ($\infty$ dilutes), $c{l}^{-}$($\infty$ diluted) and $AgC{l}_{\left(s\right)}$ are $105.579,-207.36,-167.159$ and $-127.068$ respectively. Calculate the enthalpy change for the reaction $AgN{O}_3(aq.)+HC{l}_{(aq.)}\to AgC{l}_{\left(5\right)}+HN{O}_3(aq.)$

• A. $21.471$ KJ/mol
• B. $145.688$ KJ/mol
• C. $65.488$ KJ/mol
• D. None

Answer 1

Answer: (C)

$65.488$ KJ/mol

ions	$△H_f$
$A{g}^{+}(aq.)$	$105.579$ KJ/mol
$N{O}_3^{-}(aq.)$	$-207.36$KJ/mol
$C{L}^{-}\left(aq\right)$	$-167.159$KJ/mol
$AgCL\left(s\right)$	$-129.060$KJ/mol

$A{g}^{+}(aq.)+N{O}_3^{-}\left(aq\right)\to AgN{O}_3(aq.)$

$△HfAgN{O}_3=-207.36$

$+105.579$

$-101.781$KJ/mol

$△H_f$ of $H^{+}=0$ (we know)

$H^{+}(aq.)+C{l}^{-}(aq.)=HCl(aq.)$

$△H_{f}HCl=-167.159$KJ/mol

$H^{+}+N{O}_3^{-}=HN{O}_3(aq.)$

$△H_{f}HN{O}_3=-207.36KJ/mol$

$AgN{O}_3(aq.)+HCL(aq.)\to AgCl\left(s\right)+HN{O}_3(aq.)$

$-101.781-167.159-127.68-207.36$

$△H=-{△}^o{H}_{f\left(reactant\right)}$

$=-(-127.068-207.36)+(101.781-167.159)$

$=65.488$KJ/mol