## Chemistry 9th Edition

Published by Cengage Learning

# Chapter 6 - Thermochemistry - Exercises - Page 289: 74

#### Answer

$-623 \text{ kJ}$

#### Work Step by Step

We combine the following reactions: $3N_2H_{4(l)}+3H_2O_{(l)}\Rightarrow 3N_2O_{(g)}+9H_{2(g)}\text{ 951 kJ}$ $N_2H_{4(l)}+H_2O_{(l)}\Rightarrow 2NH_{3(g)}+\frac{1}{2}O_{(g)}\text{ 143 kJ}$ $2NH_{3(g)}+3N_2O_{(g)}\Rightarrow 4N_{2(g)}+3H_2O_{(l)}\text{ -1010. kJ}$ $H_{2(g)}+\frac{1}{2}O_{2(g)}\Rightarrow H_2O_{(l)}\text{ -286 kJ}$ $8H_{2(g)}+4O_{2(g)}\Rightarrow 8H_2O_{(l)}\text{ -2288 kJ}$ To find: $4N_2H_4+4O_2\Rightarrow 4N_2+8H_2O\quad\Delta H=-2490. kJ$ Dividing $\Delta H$ by 4 to achieve the proper coefficients, we find $\Delta H = -2490./4=-623 \text{ kJ}$

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