A) CaCO3(s)CaO(s)+CO2(g) Given the following data: Temperature (K) K 298 1.93×10

Question

A) CaCO3(s)CaO(s)+CO2(g) Given the following data: Temperature (K) K 298 1.93×1023 1200 1.01 what can be said about this reaction? It will accomplish the following reaction: Given the following data: Temperature (K) intial =298 final= 1200 K initial= 1.93 x10^-23 K final=1.01 what can be said about this reaction? a) Lower temperatures result in more lime formation. b)The reaction makes more lime at higher temperatures. c)The reaction goes to completion at 1200 K. d) The equilibrium lies far to the right at room temperature. Part B For the reaction 2CO2(g)2CO(g)+O2(g), K= 1.00×1013 at 1200 K what can be said about this reaction at this temperature? For the reaction , at 1200 what can be said about this reaction at this temperature? a) The equilibrium lies far to the right. b)The reaction will proceed very slowly. c) The reaction contains significant amounts of products and reactants at equilibrium. d) The equilibrium lies far to the left. Part C For the reaction HCONH2(g)NH3(g)+CO(g), K= 4.84 at 400 K what can be said about this reaction at this temperature? For the reaction , at 400 what can be said about this reaction at this temperature? a) The equilibrium lies far to the right. b)The reaction will proceed very slowly. c)The reaction contains significant amounts of products and reactants at equilibrium. d)The equilibrium lies far to the left.

Explanation / Answer

SOLUTION:

Q1. CaCO3 <-----> CaO + CO2

At Temperature 298K K = 1.93 X 10-23 At 1200KK, K = 1.01

(A) At lowe temperature equilibrium to the left as equilibrium constant (K) has lower value, hence less lime is formed.

(B) At higher temperature e.g. 1200K; K = 1.01 hence more lime is formed

(C) At 1200K K = 1.01

We know K = [Product]/[Reactant]

or 1.01 = [Product]/[Reactant]

It is only possible when some reactant is present in the equilibium. Hence reaction does not go to completion.

Q2. 2CO2(g)2CO(g)+O2(g), K= 1.00×1013 at 1200 K

Option (B) and (D) is correct answer. A small value of K indicates the reaction proceeds slowely and the equilibrium les far to the left.

Q3. Option (A) and (C) are correct answers.

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