Then ΔH for HCNaq+NH4OHaq → Products (in KJ)

= -q2-(-q1) = (-q1 - q2)KJ = (q1-q2) +(q1-q3) +(-q1)ΔHn = q1 - q2 - q3= [-q2 -q3]+q1 KJ

Then ΔH for HCNaq+NH4OHaq → Products (in KJ)

= -q2-(-q1) = (-q1 - q2)KJ = (q1-q2) +(q1-q3) +(-q1)ΔHn = q1 - q2 - q3= [-q2 -q3]+q1 KJ

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$\large H_{aq}^ + \; + \;OH_{aq}^ - \; \to \;{H_2}{O_{(l)}},\Delta H = \; - {q_1}KJ\$
$\large HC{N_{aq}} + \;KO{H_{aq}}\; \to \;KC{N_{aq}}\; + \;{H_2}{O_{(l)}};\;\Delta H\; = \;-\;{{\text{q}}_2}{\text{ KJ}}\$
$\large HC{l_{aq}} + N{H_4}O{H_{aq}} \to N{H_4}C{l_{aq}} + {H_2}{O_{(l)}};\;\Delta H\; = \;{\text{ - }}{{\text{q}}_3}{\text{ KJ}}\$
Then ΔH for HCN_aq+NH₄OH_aq → Products (in KJ)

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[– q₂ – q₃] + q₁

– q₁ – [– q₂ – q₃]

q₁ + q₂ + q₃

– q₁ – q₂ – q₃

answer is B.

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Detailed Solution

\Delta {H_n }\; = \;\Delta {H_i}\; + \;\Delta {H_{\left( {{H^ + } + O{H^ - } \to {H_2}O} \right)}}\

{(\Delta {H_i})_{HCN}}\; = \;\Delta {H_n }\; - \;\Delta {H_{\left( {{H^ + } + O{H^ - } \to {H_2}O} \right)}}

= -q₂-(-q₁) = (-q₁ - q₂)KJ

{(\Delta {H_i})_{N{H_4}OH}}\; = \; - {q_3}\; - ( - {q_1}) = ({q_1} - {q_3})KJ

\Delta {H_n }\; \;for\;HCN\; + \;N{H_4}OH\; \to \;products\;is\

\Delta {H_n }\; = \;{(\Delta {H_i})_{HCN}}\; + \;{(\Delta {H_i})_{N{H_4}OH}}\; + \;\Delta {H_{({H^ + } + O{H^ - } \to {H_2}O)}}\

= (q₁-q₂) +(q₁-q₃) +(-q₁)
ΔH_n = q₁ - q₂- q₃
= [-q₂-q₃]+q_{1 KJ}

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Then ΔH for HCNaq+NH4OHaq → Products (in KJ)

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