∆g for these initial partial pressures is 10,403.31 KJ.
ΔG gets increasingly positive as a product gas's partial pressure is raised. ΔG becomes more negative as the partial pressure of a reactant gas increases.
∆g = RT ln (q/k)
In this equation: R = 8.314 J mol⁻¹ K⁻¹ or 0.008314 kJ mol⁻¹ K⁻¹
K = 325
If ΔG < 0, then K > Q, and the reaction must proceed to the right to reach equilibrium.
∴∆g = RT ln (q/k)
= 8.314 × 298 ln ( 5 / 325)
= 2477.57 ln 0.015
= 2477.57 × (-4.199)
= 10,403.31 KJ
Products are preferred over reactants at equilibrium if G° 0 and both the products and reactants are in their standard states. When reactants are preferred above products in equilibrium, however, if G° > 0, K 1. At equilibrium, neither reactants nor products are preferred if G° = 0, hence K = 1.
Therefore, ∆g for these initial partial pressures is 10,403.31 KJ.
Learn more about equilibrium here:
brainly.com/question/13414142
#SPJ4
Some of the scientific questions that may be answered through the experiment are:
(1) What are the physical changes that may occur in the solution or the indicator when added with acidic/basic solution?
(2) How much of the indicator is needed in order to bring about a significant physical change in the solution to identify its H+ concentration?
For the most part aesthetics but you could argue they reduce air turbulence in some cases which would lead to better gas mileage. On sports cars however the goal is usually down force which is not wanted on economy cars as it can actually worsen gas mileage.
I hope that answered your question.
Answer:
A , power
Explanation:
Hope this is useful. Have a lovely rest of your day! God bless you.
