If Kc = 0.460 at 40.°C and Kc = 0.685 at 90.°C, what is ΔH° for the reaction?
If Kc = 0.440 at 40.°C and Kc = 0.515 at 90.°C, what is ΔH° for the reaction? X <---> Y
1. If If Kc=0.425 at 40.°C and Kc=0.670 at 90.°C, what is ΔH° for the reaction? X↽−−⇀Y kj 2.Consider a general reaction A(aq)⇌enzymeB(aq) The ΔG°′ of the reaction is −7.500 kJ·mol−1. Calculate the equilibrium constant for the reaction at 25 °C. Keq′= What is ΔG for the reaction at body temperature (37.0 °C) if the concentration of A is 1.7 M and the concentration of B is 0.65 M? ΔG= 3. Attempt 9 Consider the data in the table. Compound...
Automobiles and trucks pollute the air with NO. At 2000.0°C, Kc for the reaction At 2000.0°C Kc = 4.100 × 10–4 and ΔH° = 180.6 kJ. What is the value of Kc at 69.00°C?
Consider the following reaction at 298 K. C(graphite)+2H2(g)⟶CH4(g)ΔH∘=−74.6 kJ and ΔS∘=−80.8 J/KC(graphite)+2H2(g)⟶CH4(g)ΔH∘=−74.6 kJ and ΔS∘=−80.8 J/K Calculate the following quantities. ΔSsys=ΔSsys J/K ΔSsurr= J/K ΔSuniv= J/K
Automobiles and trucks pollute the air with NO. At 2000.0°C, Kc for the reaction $$N2(g)+O2(g)2NO(g) At 2000.0°C Kc = 4.100 × 10–4 and ΔH° = 180.6 kJ. 1st attempt What is the value of Kc at 43.00°C?
2Na(s)+Cl2(g) ⟶2NaCl(s)+819 kJWhat is the ΔH for the above reaction? Is the reaction endothermic or exothermic? How many joules (J) of heat are released when 0.460 g of sodium reacts with chlorine gas?
A-><-B The value of Kc for the given generic reaction is 0.445 at 50°C and 0.655 at 100°C. 1st attempt See Periodic Table Calculate ΔH° for the reaction.
If K_c = 0.450 at 40.degree C and K_c = 0.530 at 90.degree C, what is Delta H degree for the reaction? X Y
You are studying the reaction: A(aq) ↔ 3 B(aq) The value of Kc for this reaction is 0.685. When the reaction reaches equilibrium, you find [A]eq=0.133M. What is [B]eq? Report your answer to 3 decimal places. Your Answer:
For reactions carried out under standard-state conditions, the equation ΔG = ΔH − TΔS becomes ΔG° = H° − TΔS°. Assuming ΔH° and ΔS° are independent of temperature, one can derive the equation: ln( K2 K1 ) = ΔH° R ( T2 − T1 T1T2 ) where K1 and K2 are the equilibrium constants at T1 and T2, respectively. Given that at 25.0°C, Kc is 4.63×10−3 for the reaction N2O4(g) longrightleftarrow 2NO2(g) ΔH° = 58.0 kJ/mol calculate the equilibrium constant...