First slide

Hess law of constant summation

Question

ΔH_f of BaCO_3(s), CO_2(g) , HCl_(aq) and H₂O_(l) respectively are –1216, – 393, – 176 and – 286 KJ/mol. Then ΔH_f of BaCl_2(aq) from BaCO_3(s)+ 2HCl_(aq) → BaCl_2(aq)+ CO_2(g)+ H₂O_(l), ΔH = – QKJ

Easy

Solution

$\large {\Delta _r}H\; = \;\sum {{\Delta _p}H\; - \;\sum {{\Delta _R}H} }$
$\large - Q = {\Delta _f}{H_{(BaC{l_2})}} + ( - 393) + ( - 286) - [ - 1216 + 2( - 176)]$
$\large - Q\; = \;{\Delta _f}{H_{(BaC{l_2})}}\; + \;889\$
$\large {\Delta _f}{H_{(BaC{l_2})}}\; = \; - (889 + Q)KJ$

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