Answer:
D - chemical and physical
Explanation:
Only chemical and physical changes can reach the level of dynamic equilibrium. Nuclear reactions cannot reach dynamic equilibrium.
- A system is in dynamic equilibrium when the rate of forward reaction is the same as that of backward reaction in a reversible reaction.
- Nuclear reactions cannot be reversed.
- Dynamic equilibrium is prominent in chemical reactions. It is commonly found that as a reaction occurs, the backward and forward reactions can reach equilibrium levels.
- In physical changes, this can also occur when certain conditions of pressure and temperatures are satisfied.
Answer: In gases the particles move rapidly in all directions, frequently colliding with each other and the side of the container. With an increase in temperature, the particles gain kinetic energy and move faster. The actual average speed of the particles depends on their mass as well as the temperature – heavier particles move more slowly than lighter ones at the same temperature. The oxygen and nitrogen molecules in air at normal room temperature are moving rapidly at between 300 to 400 metres per second. Unlike collisions between macroscopic objects, collisions between particles are perfectly elastic with no loss of kinetic energy.
Explanation: This is very different to most other collisions where some kinetic energy is transformed into other forms such as heat and sound. It is the perfectly elastic nature of the collisions that enables the gas particles to continue rebounding after each collision with no loss of speed. Particles are still subject to gravity and hit the bottom of a container with greater force than the top, and giving gases weight. Hope this helps with your problem! Byeeee :DDD
Answer:
1) ΔG°r(298 K) = - 28.619 KJ/mol
2) ΔG°r will decrease with decreasing temperature
Explanation:
- CO(g) + H2O(g) → H2(g) + CO2(g)
1) ΔG°r = ∑νiΔG°f,i
⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)
from literature, T = 298 K:
∴ ΔG°CO2(g) = - 394.359 KJ/mol
∴ ΔG°CO(g) = - 137.152 KJ/mol
∴ ΔG°H2(g) = 0 KJ/mol........pure substance
∴ ΔG°H2O(g) = - 228.588 KJ/mol
⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )
⇒ ΔG°r(298 K) = - 28.619 KJ/mol
2) K = e∧(-ΔG°/RT)
∴ R = 8.314 E-3 KJ/K.mol
∴ T = 298 K
⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6
⇒ ΔG°r = - RTLnK
If T (↓) ⇒ ΔG°r (↓)
assuming T = 200 K
⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)
⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol
