When we have the balanced reaction equation is:
H2(g) + CO2(g) ↔ H2O(g) + CO (g)
a) first, to calculate ΔG° for the reaction:
we will use this formula:
ΔG° = -RT㏑Kp
when R is R- rydberg constant = 8.314J/mol.K
and T is the temperature in Kelvin = 2000 K
and Kp = 4.4
so, by substitution:
ΔG° = - 8.314 *2000 *㏑4.4
= - 24624 J/mol = - 24.6 KJ/mol
b) to calculate ΔG so, we will use this formula:
ΔG = ΔG° + RT㏑Qp
So we need first, to get Qp from the reaction equation:
when Qp = P products / P reactants
= PH2O*PCO / PH2 * PCO2
= (0.66 atm * 1.2 atm) / (0.25 * 0.78)
= 4.1
so by substitution:
ΔG = -24624 + 8.314*2000*㏑4.1
= -1162 J/mol = - 1.16 KJ/mol
Answer:
Stars shine because they are extermly hot of light .The source of their energy is niclear reactions going on deep inside the stars.
Explanation:
Answer:
The enthalpy for given reaction is 232 kilo Joules.
Explanation:
...[1]
..[2]
..[3]
..[4]
2 × [2] + [3] - [1] ( Using Hess's law)
The enthalpy for given reaction is 232 kilo Joules.
from ICE table
H2(g) + I2 (g )↔ 2HI(g)
equ 0.958 0.877 0.02 first mix1
0.621 0.621 0.101 sec mix2
Kp1 = P(HI)^2 / p(H2)*p(I2) for mix 1
= 0.02^2 / 0.958*0.877
= 4.8x10^-4
Kp2 = P(HI)^2 / P(H2)* P(I2) for mix 2
= 0.101^2/ 0.621*0.621
= 0.0265
we can see that Kp1< Kp2 that means that the sec mixture is not at equilibrium. It will go left to reduce its products and increase reactant to reduce the Kp value to achieve equilibrium.
and the partial pressure of Hi when mix 2 reach equilibrium is:
4.8x10^-4 = P(Hi)^2 / (0.621*0.621)
∴ P(Hi) at equilibrium = 0.0136 atm
