Determine δh for reaction 4 from these data: reaction 4: n2h4 (l) + 2h2o2 (l) → n2 (g) + 4h2o (l) δh 4 = ? reaction 1: n2h4 (l) + o2 (g) → n2 (g) + 2h2o (l) δh 1 = -622.3 kj reaction 2: h2 (g) + 1/2o2 (g) → h2o (l) δh 2 = -285.8 kj reaction 3: h2 (g) + o2 (g) → h2o2 (l) δh 3 = -187.8kj

2 Answers

Anonymous

7 years ago

Best Answer

Q = m*c*ΔT

Applying the first law of thermodynamics or the law of conservation of energy i.e  

Energy lost by the metal == Energy gained by the water  so now:

m1 * c1 *( Tinitial - Tfinal ) ==  m2 * c2 *( Tfinal- Tinitial )  

(2.4)*(c1)*(98.88-20.60) == (28)*(4.2)*(20.60 - 19.73)

c1 = the specific heat of the unknown metal

c2 =the specific heat of water 4.2J/g K

Solve for c1

c1* 187.872 == 102.312

c1 == 0.5446 J/g ▫C

Note this:  

1. Changing the temperature scale from Celsius to Kelvin is quite irrelevant here, as we are working with the difference only.

2. The interchanging of ( Tinitial - Tfinal) to *( Tfinal- Tinitial ), because of the substance is losing energy, yet while the other is gaining.

Explanation:

hope i helped : )

the product of the pressure and volume is also constant

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