Question

Consider the following process $\Delta$ $H\left(kJ/mol\right)$
$\frac{1}{2}$ $A\to B$ $+150$
$3B$ $⟶$ $2C+D$ $-125$
$E+A$ $⟶$ $2D$ $+350$
For; $B+D$ $⟶$ $E+2C$ , $\Delta H$ will be :

Answer 1

Given: $\frac{1}{2}A\to B$....(i) $△H=150KJ/mol$

$3B\to 2C+D$....(ii) $△H=-125KJ/mol$

$E+A\to 2D$....(iii) $△H+350KJ/mol$

To find - $B+D\to E+2C$ $△H=?$

Subtract equation (iii) from (i) and multiply reaction (i) by (ii)

$A\to 2B$ $△H=150\times 2KJ/mol$....(v)

$3B\to 2C+D$ $△H=-125KJ/mol$

$E+A\to 2D$ $△H=300KJ/mol$

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$3B-E-A\to 2C-D$ $△H=-475KJ/mol$

$3B+D\to E+A+2C$....(iv) $△H=-475KJ/mol$

From reaction (v), by reversing it

$2B\to A$ $△H=-300KJ/mol$....(vi)

From reaction (v) and (vi)

$3B+D\to E+A+2C$ $△H=-475KJ/mol$

$2B\to A$ $△H=-308KJ/mol$

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$B+D\to E+2C$ $△H=-175KJ/mol$