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

Question

contestada

Respuesta :

ummyesmam ummyesmam · Answer 1 · 2019-10-01T16:21:05+02:00

Answer:

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 : )