Question

Consider the following processes :
$\Delta H$ (kJ/mol)
$1/2 A \rightarrow B$ - +150
$3B \rightarrow 2C + D$ - –125
$E + A \rightarrow 2D$ - +350
For $B + D \rightarrow E + 2C$, $\Delta H$ will be :
(a) 525 kJ/mol
(b) – 175 kJ/mol
(c) – 325 kJ/mol
(d) 325 kJ/mol

Answer 1

Answer:

(c) – 325 kJ/mol

Is your Correct Options !

Thanks ❤️

Answer 2

Answer:

-175

Explanation:

(1/2A= B )*2. 150 KJ/MOL

A= 2B. 300. KJ/MOL

3B =2C+D. -125 KJ/MOL

2D = A+E. -350 KJ/MOL

B+D = A+2C

∆H = (300-125-350)

∆H = -175 KJ/MOL