Hydrazine, a component of rocket fuel, undergoes combustion to yield $N_{2}$ and $H_{2} O .$
$N_{2} H_{4} (l) + O_{2} (g) \to N_{2} (g) + 2 H_{2} O (l)$
What is the change in internal energy for above reaction of $N_{2} H_{4}$ (in kJ/mole)?
Given: Reaction $Δ H (k J)$
$2 N H_{3} (g) + 3 N_{2} O (g) \to 4 N_{2} (g) + 3 H_{2} O (l)$    $- 1011 k J$
$N_{2} O (g) + 3 H_{2} (g) \to N_{2} H_{4} (l) + H_{2} O (l)$    $- 317 k J$
$4 N H_{3} (g) + O_{2} (g) \to 2 N_{2} H_{4} (l) + 2 H_{2} O (l)$ $- 286 k J$
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$    $- 285 k J$

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$- 622.75$

None of these

$- 620.5$

$1167.5$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$0.5 N H_{3} + 0.75 N_{2} O \to N_{2} (g) + 0.75 H_{2} O (l) - - - - (i) 0.75 N_{2} H_{4} + 0.75 H_{2} O \to 0.75 N_{2} O (g) + 2.25 H_{2} (g) - - - (i i) 0.25 N_{2} H_{4} + 0.25 H_{2} O \to 0.5 N H_{3} + \frac{1}{8} O_{2} (g) - - - - - (i i i) 2.25 H_{2} + 1.125 O_{2} \to 2.25 H_{2} O (g) - - - - - - (i v) \frac{1}{4} (i) - \frac{3}{4} (i i) - \frac{1}{8} (i i i) + \frac{9}{4} (i v) = - 620.5$