Data Table 2: A. Use your results to determine if the forward reaction in the po
ID: 1009986 • Letter: D
Question
Data Table 2:
A. Use your results to determine if the forward reaction in the potassium chromate/HCl reaction endothermic or exothermic. Explain your answer, using Table 1 to help construct your thoughts.
B. Write the equation for the equilibrium constant (K) of the reaction studied in this exercise.
2 K2CrO4 + 2HCl Cr2O7 + H2O + 2KCl
Chromate Dichromate
Use the information below to answer Questions 3, 4, and 5:
The equilibrium constant (K) of the reaction below is K = 6.0 x 10-2, with initial concentrations as follows: [H2] = 1.0 x 10-2 M, [N2] = 4.0 M, and [NH3] = 1.0 x 10-4M.
C. If the concentration of the reactant H2 was increased from 1.0 x 10-2 M to 2.5 x 10-1M, calculate the reaction quotient (Q) and determine which way the chemical system would shift.
D. If the concentration of the reactant H2 was decreased from 1.0 x 10-2 M to 2.7 x 10-4M, calculate the reaction quotient (Q) and determine which way the chemical system would shift.
E. If the concentration of the product NH3 was increased from 1.0 x 10-4 M to 5.6 x 10-3M, calculate the reaction quotient (Q) and determine which way the chemical system would shift.
Im having a hard time understanding this.
Color of Chromate Color of Dichromate Number of drops of NaOH to shift chemical system yellow Orange 5Explanation / Answer
(A): Since the reaction proceeds towards the right(product side) under hot water bath (higher temperature), the reaction needs heat to proceed towards right. Hence the reaction is endothermic reaction.
(B): Equilibrium constant is written on the basis of the net ionic equation. The net ionic equation is
2CrO42-(aq) + 2H+(aq) ------- > Cr2O72-(aq) + H2O
Hence equilibrium constant K = [Cr2O72-(aq)] / ([CrO42-(aq)]2x[H+(aq)]2) (answer)
(C): Given the new concentration of H2, [H2]' = 2.5x10-1 M
For the given reaction, reaction quotient, Q = [NH3(g)]2 / [N2(g)]x[H2(g)]3
=> Q = (1.0x10-4 M)2 / (4.0 M)x(2.5x10-1 M)3
=> Q = 1.6x10-7 (answer)
Since Q < K, to increase the value to Q the reaction will move towards product side (answer).
(D):
Given the new concentration of H2, [H2]' = 2.7x10-4 M
For the given reaction, reaction quotient, Q = [NH3(g)]2 / [N2(g)]x[H2(g)]3
=> Q = (1.0x10-4 M)2 / (4.0 M)x(2.7x10-4 M)3
=> Q = 1.27x102 (answer)
Since Q > K, to decrease the value to Q the reaction will move towards reactant side (answer).
(E):
Given the new concentration of H2, [H2]' = 5.6x10-3 M
For the given reaction, reaction quotient, Q = [NH3(g)]2 / [N2(g)]x[H2(g)]3
=> Q = (1.0x10-4 M)2 / (4.0 M)x(5.6x10-3 M)3
=> Q = 1.423x10-2 (answer)
Since Q < K, to increase the value to Q the reaction will move towards product side (answer).
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