Question

$A+B\to C△H=-390kJ$
$D+\frac{1}{2}B\to E△H=-280kJ$
$F+\frac{1}{2}B\to C+E△H=-275kJ$
From the above data given, find the value for $△H$ for the reaction: $D+A+B\to F$?

• A. $-165kJ$
• B. $+385kJ$
• C. $-395kJ$
• D. $-945kJ$
• E. $+400kJ$

Answer 1

Answer: (C)

$-395kJ$

$A+B\to C$ $\Delta H=-390kJ$ ---------------1

$D+\frac{1}{2}B\to E$ $\Delta H=-280kJ$ ------------2

$F+\frac{1}{2}B\to C+E$ $\Delta H=-275kJ$ ---------3

By Adding Eq 1, 2 and subtracting Eq 3, we get

$D+A+B\to F$

So, $\Delta H=(-390)+(-280)-(-275)kJ$

$\Delta H=-395kJ$