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
Answer:
decrease the volume of the cylinder.
Explanation:
In order to be able to solve this question we have to understand what Boyle's law is. According to Boyle's law; at constant temperature the pressure of a given mass of gas is inversely proportional to to its volume.
The Boyle's law shows us the relationship between the pressure and the volume. So, the Important thing to note hear is that if the volume in a container is decreased then the pressure will increase (and vice versa) due to the fact that as the volume decreases the particles in that container makes more collision which will make the pressure to increase.
Since, the piston is moveable it means that we can decrease and increase the volume in the cylinder. So, if the decrease the volume of the cylinder then we will have an increase in the pressure of the gas below the piston.
The answer would be 2+ since the atomic number represents how many protons are in the element. In this case, there are 16 protons, but only 14 electrons, which means there are an additional 2 protons, hence the 2+ charge on the ion.
The question asks average kinetic energy. So it is only related with the temperature. The higher temperature is, the higher kinetic energy is. So the answer is (4).
