Answer to Question #141815 in General Chemistry for Christian

Question

Hydrazine, N2H4(l)is used as a rocket fuel. The thermochemical equation for the combustion of hydrazine is: N2H4+ O2→ N2+ 2 H2O ∆H = -622.4 kJ. What quantity of heat is liberated by the combustion of 1.00 g of hydrazine?

Answer 1

Q141815,

Hydrazine, N2H4(l)is used as a rocket fuel. The thermochemical equation for the combustion of hydrazine is: N2H4+ O2→ N2+ 2 H2O ∆H = -622.4 kJ. What quantity of heat is liberated by the combustion of 1.00 g of hydrazine?

Solution:

The equation for the combustion of hydrazine is

N₂H₄ + O₂ -----> N₂ + 2H₂O; ∆H = -622.4 kJ.

In this equation, there is no coefficient given to N₂H_{4 ,} O₂, and N₂ .

This means their coefficient is 1.

1 N₂H₄ + 1 O₂ -----> 1 N₂ + 2H₂O; ∆H = -622.4 kJ.

A negative sign of ∆H indicates that heat is given out in the given reaction.

The coefficient of N₂H₄ is 1, so this means 1 mole of N₂H₄ on combustion gives 622.4 kJ of heat.

Step 1 : Convert 1.00g of Hydrazine, N₂H ₄ (l) to moles using the molar mass of Hydrazine, N₂H₄ (l)

molar mass of N₂H₄ = 2 * atomic mass of N + 4 * atomic mass of H

= 2 * 14.007g/mol + 4 * 1.00794 g/mol

= 28.014g/mol + 4.03176g/mol

= 32.0456g/mol

Using molar mass 32.0456g/mol convert 1.00g of N₂H₄ to moles.

Moles of N₂H₄ = 1.00g N₂H₄ * 1 mol N₂H₄ / 32.0456g N₂H₄

= 0.031206 mol of N₂H₄ .

Step 2: Using, 1mol N₂H₄ = 622.4 kJ ,find the amount of heat given out by 0.031206 mol of N₂H₄ .

amount of heat released by burning 1.00g N₂H₄

= 0.031206 mol of N₂H₄ * 622.4 kJ /1 mol of N₂H₄

= 19.422 kJ.

In question, we are given 1.00g hydrazine which is in 3 significant figures, so our final answer must also be in 3 significant figures.

Hence amount of heat released by burning 1.00g hydrazine (N₂H₄ ) is 19.4 kJ