For a particular reaction at 235.8 °C, ΔG=−936.92 kJ/mol , and ΔS=513.79 J/(mol⋅K) . Calculate ΔG for this reaction at −9.9 °C.

-138.9 kJ/mol

Explanation:

Step 1: Convert 235.8°C to the Kelvin scale

We will use the following expression.

K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K

Step 2: Calculate the standard enthalpy of reaction (ΔH°)

We will use the following expression.

ΔG° = ΔH° - T.ΔS°

ΔH° = ΔG° / T.ΔS°

ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K

ΔH° = -3.583 kJ (for 1 mole of balanced reaction)

Step 3: Convert -9.9°C to the Kelvin scale

K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K

Step 4: Calculate ΔG° at 263.3 K

ΔG° = ΔH° - T.ΔS°

ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K

ΔG° = -138.9 kJ/mol

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