A asystem at equilibrium stops
Answer:
90.3 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
2 NO(g) + O₂(g) → 2 NO₂(g) ∆H°rxn = –114.2 kJ
We can find the standard enthalpy of formation for NO using the following expression.
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))
ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol
ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol
ΔH°f(NO(g)) = 90.3 kJ/mol
<span>Chromium is a transition metal and it has 24 electrons and here is the orbital diagram. If we're going to make this short hand and make the electron configurationfor this we would make this 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d4 okay from now on every time you see 3d4 you're going to change it, we do not like 3d4.</span>
Answer:
Reverse the 
reaction
Explanation:
Reactions:
Overall:
As can be seen, in the overall reaction we have 
in the reactants like in the second reaction and 
in the products. The is in the first reaction but as a reactant so we need to reverse that reaction:
Combining:
