Question 1315 mol of AB(g) is placed in an empty 1,0 dmcontainer. Equilibrium is reached at 500 K after6 minutes.2 AB(g)A2(g) + B2(g)There is 5,0 mol B2(g) in the reaction mixture atequilibrium.13.1 Calculate:13.1.1 the concentration of each substanceat equilibrium. 13.1.2 the equilibrium constant (Kc) at 500 K.13.2 The temperature is increased to 800 K after12 minutes. The forward reaction is favouredmore than the reverse reaction. A newequilibrium is reached after 18 minutes.13.2.1 Draw a graph of reaction rate againsttime on which you show all changes tothe rate over the 18 minutes.13.2.2 Is the forward reaction exothermic orendothermic? Explain.13.2.3 How will the Ke value change at thenew temperature of 800 K?

formula used :- Kc = [product]/[Reactant]

Question 13 15 mol of AB(g) is placed in an empty 1,0 dm container. Equilibrium is reached at 500 K after 6 minutes. 2 AB(g) = A2(g) + B2(g) There is 5,0 mol B2(g) in the reaction mixture at equilibrium. 13.1 Calculate: 13.1.1 the concentration of each substance at equilibrium.

Question 13 15 mol of AB(g) is placed in an empty 1,0 dm container. Equilibrium is reached at 500 K after 6 minutes. 2 AB(g) = A2(g) + B2(g) There is 5,0 mol B2(g) in the reaction mixture at equilibrium. 13.1 Calculate: 13.1.1 the concentration of each substance at equilibrium.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

13.1.2 the equilibrium constant (Kc) at 500 K.
13.2 The temperature is increased to 800 K after
12 minutes. The forward reaction is favoured
more than the reverse reaction. A new
equilibrium is reached after 18 minutes.
13.2.1 Draw a graph of reaction rate against
time on which you show all changes to
the rate over the 18 minutes.
13.2.2 Is the forward reaction exothermic or
endothermic? Explain.
13.2.3 How will the Ke value change at the
new temperature of 800 K?

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