For the reaction shown below, K_eq = 2 Times 10^12. Which of the following state

Question

For the reaction shown below, K_eq = 2 Times 10^12. Which of the following statements concerning this system all equilibrium is true? 2CO(g) + O_2(g) 2CO_2(g) The equilibrium lies to the left. The equilibrium solution contains equal amounts of CO, O_2, and CO_2. The reaction is very fast, due to the high value fof K_eq. The equilibrium solution contains predominantly CO_2. The equilibrium system contains almost twice as many reactant molecules as product molecules. A carbohydrate sample weighing 0.235 g was found to have a feel value of 3.84 kJ. What is the fuel value of one gram of this carbohydrate, in nutritional Calories? 3,910 Cal 0.535 Cal 16.3 Cal Cal 3.91 Cal The reaction A + 2B + C rightarrow D + 2E is first order n reactant A, first order in B, and second order in C. What is the rate law equation for this reaction? Which of the following is the equilibrium constant expression for the reaction? 2NO(g) + O_2(g) rightarrow 2NO_2(g) In the industrial synthesis of ammonia, the equilibrium constant expression may be written as: K_eq = [NH_3]^2/[N_2][H_2]^3 Calculate the value of this equilibrium constant, if the equilibrium concentration of nitrogen in the reaction mixture at 600 degree C if [N_2] = 4.53 M; [H_2] = 2.49 M; and [NH_3] = 7.62 M 5.15 2.07 0.830 0.676 1.44 When a sample of aqueous hydrochloric acid was neutralized with aquous sodium hydroxide in a calorimeter, the temperature of 100.0 g of water surrounding the reaction increased from 25.0 degree C to 31.5 degree C. If the specific heat of water is 1.00 cal/(g degree C), calculate the quantity of energy in calories involved in this neutralization reaction. 1000 cal 100.0 cal 6.50 cal 1250 cal 650 cal

Explanation / Answer

9) at equilibrium the rate of forward reaction becomes equal to rate of backward reaction

so at equilibrium there will some amount of products as well as reactants

However as the value of rate constant is very high, there will be predominantly CO2.

10) The heat evolved due to complete combustion of 0.235grams of fuel = 3.84 KJ = 3.84 / 4.184 Kcalories = 0.918 Kcal

So the heat evolved by 1 gram of fuel = 0.918 Kcal / 0.235 = 3.91 Kcal [the nutritional calories]

11) The rate law will be

Rate = K [A][B][C]2

12) The Keq = Product of concentration of products / product of concentration of reactants

Keq = [NO2]2 / [NO]2[O2]

13) Keq = [NH3]2 / [N2][H2]3 = [7.62]2 / [4.53] [ 2.49]3 = 0.830

14) The heat absorbed by water= heat released in the reaction

Heat absorbed by water = Mass X specific heat of water X change in temperature

Heat absorbed by water= 100 X 1 X (31.5-25) = 650 Calories

This is the heat evolved during the reaction

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