Areaction that occurs in the internal combustion engine is n2(g) + o2(g) ⇌ 2 no(g) (a) calculate δh o and δs o for the reaction at 298 k. δh o = kj δs o = j/k (b) assuming that these values are relatively independent of temperature, calculate δg o at 55°c, 2570°c, and 3610°c. 55°c: kj 2570°c: kj 3610°c: kj

Explanation:1) ΔrH = 2mol·ΔfH(NO) - (ΔfH(O₂) + ΔfH(N₂)).

ΔrH = 2 mol · 90.3 kJ/mol - (0 kJ/mol + 0 kJ/mol).

ΔrH = 180.6 kJ.

2) ΔS = 2mol·ΔS(NO) - (ΔS(O₂) + ΔS(N₂)).

ΔS = 2mol · 210.65 J/mol·K - (1mol · 205 J/mol·K + 1 mol · 191.5 J/K·mol).

ΔS = 24.8 J/K.

3) ΔG = ΔH - TΔS.

55°C: ΔG = 180.6 kJ - 328.15 K · 24.8 J/K = 172.46 kJ.

2570°C: ΔG = 180.6 kJ - 2843.15 K · 24.8 J/K = 110.09 kJ.

3610°C: ΔG = 180.6 kJ - 3883.15 K · 24.8 J/K = 84.29 kJ.

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

the product of the pressure and volume is also constant