Answer to Question #149846 in General Chemistry for taiya

Question

The following reaction was allowed to go to completion and produced a total of 30.4 mg NH3. If the energy transfers into 350 grams of air and increases the temperature only 4⁰C, what is the specific heat of air?
N2 + 3H2 --> 2NH3 ∆H = - 92.6 kJ /mol

Answer 1

$N2 + 3H2 \to 2NH3\\ ∆H = - 92.6 kJ /mol$

From the reaction above, when 2 moles of NH3 is produced, 92.6kJ is evolved

This means that 34g of NH3 = 92.6kJ

30.4mg = xkJ

x = (30.4/1000g × 92.6)/34g = 0.0828kJ = 82.8J.

This means 82.8J is evolved in the reaction.

Using, H = mc∆T

H = 82.8J

m = 350g

∆T = 4

c = H/m∆T = 82.8/350×4 = 0.0591 J/g.°C