Home
Class 12
CHEMISTRY
An equilibrium mixture in a vessel of ca...

An equilibrium mixture in a vessel of capacity 100 litre contain 1 mol `N_(2)`, 2 mol `O_(2)` and 3 mol NO. Find number of moles of `O_(2)` to be added, so that at new equilibrium the concentration of No is 0.04 mol/lit.

A

`(101)/(18)`

B

`(101)/(9)`

C

`(202)/(9)`

D

None of these

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will follow the equilibrium process and calculations systematically. ### Step 1: Write the balanced chemical equation The balanced equation for the reaction is: \[ N_2 + O_2 \rightleftharpoons 2 NO \] ### Step 2: Determine initial moles and concentrations Given: - Moles of \( N_2 = 1 \) mol - Moles of \( O_2 = 2 \) mol - Moles of \( NO = 3 \) mol - Volume of the vessel = 100 L To find the initial concentrations: - Concentration of \( N_2 = \frac{1 \text{ mol}}{100 \text{ L}} = 0.01 \text{ mol/L} \) - Concentration of \( O_2 = \frac{2 \text{ mol}}{100 \text{ L}} = 0.02 \text{ mol/L} \) - Concentration of \( NO = \frac{3 \text{ mol}}{100 \text{ L}} = 0.03 \text{ mol/L} \) ### Step 3: Calculate the equilibrium constant \( K_c \) Using the equilibrium expression: \[ K_c = \frac{[NO]^2}{[N_2][O_2]} \] Substituting the initial concentrations: \[ K_c = \frac{(0.03)^2}{(0.01)(0.02)} = \frac{0.0009}{0.0002} = 4.5 \] ### Step 4: Set up the new equilibrium after adding \( X \) moles of \( O_2 \) Assume \( X \) moles of \( O_2 \) are added. The new moles will be: - Moles of \( N_2 = 1 \) mol - Moles of \( O_2 = 2 + X \) mol - Moles of \( NO = 3 + 2\alpha \) mol (where \( \alpha \) is the change in moles of \( NO \)) ### Step 5: Express the new concentrations The new concentrations will be: - Concentration of \( N_2 = \frac{1}{100} \) - Concentration of \( O_2 = \frac{2 + X}{100} \) - Concentration of \( NO = \frac{3 + 2\alpha}{100} \) ### Step 6: Set the new concentration of \( NO \) We are given that the concentration of \( NO \) at new equilibrium is \( 0.04 \text{ mol/L} \): \[ \frac{3 + 2\alpha}{100} = 0.04 \] Multiplying through by 100: \[ 3 + 2\alpha = 4 \implies 2\alpha = 1 \implies \alpha = 0.5 \] ### Step 7: Substitute \( \alpha \) back into the equilibrium expression Now we substitute \( \alpha \) into the equilibrium expression: \[ K_c = \frac{(0.04)^2}{\left(\frac{1}{100}\right)\left(\frac{2 + X}{100}\right)} \] This simplifies to: \[ K_c = \frac{0.0016}{\frac{(2 + X)}{10000}} = \frac{0.0016 \times 10000}{(2 + X)} = \frac{16}{(2 + X)} \] ### Step 8: Set the equation equal to the calculated \( K_c \) Setting the two expressions for \( K_c \) equal: \[ \frac{16}{(2 + X)} = 4.5 \] ### Step 9: Solve for \( X \) Cross-multiplying gives: \[ 16 = 4.5(2 + X) \implies 16 = 9 + 4.5X \implies 7 = 4.5X \implies X = \frac{7}{4.5} \approx 1.56 \] ### Final Answer The number of moles of \( O_2 \) to be added is approximately \( 1.56 \) moles. ---
Promotional Banner

Similar Questions

Explore conceptually related problems

An equilibrium mixture in a vessel of capacity 100 litre contain 1 "mol" N_(2).2"mol"O_(2) "and" 3 "mol" NO . Number of moles of O_(2) to be added so that at new equilibrium the conc. Of NO is found to be 0.04 mol//lt.

The reaction, CO(g)+3H_(2)(g) hArr CH_(4)(g)+H_(2)O(g) is at equilibrium at 1300 K in a 1 L flask. It also contains 0.30 mol of CO, 0.10 mol of H_(2) and 0.02 mol of H_(2)O and an unknown amount of CH_(4) in the flask. Determine the concentration of CH_(4) in the mixture. The equilibrium constant K_(c ) for the reaction at the given temperature us 3.90 .

An equilibrium mixture at 300 K contains N_(2)O_(4) and NO_(2) at 0.28 and 1.1 atm , respectively. If the volume of container is doubles, calculate the new equilibrium pressure of two gases.

A reaction system in equilibrium according to reaction 2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g) in one litre vessel at a given temperature was found to be 0.12 mole each of SO_(2) and SO_(3) and 5 mole of O_(2) In another vessel of one litre contains 32 g of SO_(2) at the same temperature. What mass of O_(2) must be added to this vessel in order that at equilibrium 20% of SO_(2) is oxidized to SO_(3) ?

An equilibrium mixture CO(g)+H_(2)O(g) hArr CO_(2)(g)+H_(2)(g) present in a vessel of one litre capacity at 815^(@)C was found by analysis to contain 0.4 mol of CO, 0.3 mol of H_(2)O, 0.2 mol of CO_(2) and 0.6 mol of H_(2) . a. Calculate K_(c) b. If it is derived to increase the concentration of CO to 0.6 mol by adding CO_(2) to the vessel, how many moles must be addes into equilibrium mixture at constant temperature in order to get this change?

1 mol of N_(2) and 2 mol of H_(2) are allowed to react in a 1 dm^(3) vessel. At equilibrium, 0.8 mol of NH_(3) is formed. The concentration of H_(2) in the vessel is

What mass of oxygen will contain 2 mol of O_(2) molecules? Molar mass of O_(2) is 32 g mol^(-1) .

A mixture of 2 moles of N_(2) and 8 moles of H_(2) are heated in a 2 lit vessel, till equilibrium is established. At equilibrium, 04 moles of N_(2) was present. The equilibrium concentration of H_(2) will be

Equilibrium constants of reaction CaCO_3(s)

The reaction , N_(2)(g) + O_(2) (g) hArr 2 NO(g) contribute to air pollution whenever a fuel is burnt in air at a high temperature . At 1500 K , quilibrium constant K for its is 1.0 xx 10^(-5) . Suppose in a case [N_(2)] = 0.80 mol L^(-1) and [O_(2)] = 0.20 mol L^(-1) before any reaction occurs . Calculate the equilibrium concentrations of the reactants and the product after the mixture has been heated to 1500 K.