Answer:
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∆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.
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Study of interaction between matter and energy using mechianical description
Answer:
15.8 ft/s
Explanation:
= Velocity of car A = 9 ft/s
a = Distance car A travels = 21 ft
= Velocity of car B = 13 ft/s
b = Distance car B travels = ft
c = Distance between A and B after 4 seconds = √(a²+b²) = √(21²+28²) = √1225 ft
From Pythagoras theorem
a²+b² = c²
Now, differentiating with respect to time
∴ Rate at which distance between the cars is increasing three hours later is 15.8 ft/s
Absorption occurs when <span>all of the energy from light waves is transferred to a medium! </span>
