Home
Class 11
CHEMISTRY
The equilibrium constant for the reactio...

The equilibrium constant for the reaction `CO(g)+H_(2)O(g)hArrCO_(2)(g)+H_(2)(g) is 5` how many moles of `CO_(2)` must be added to 1 litre container alrady containing 3 moles each of CO and `H_(2)O` to make 2 M equilibrium conentration of CO?

A

15

B

19

C

5

D

20

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will follow these steps: ### Step 1: Write the balanced chemical equation The reaction given is: \[ \text{CO(g)} + \text{H}_2\text{O(g)} \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g) \] ### Step 2: Identify initial conditions We are given that initially, there are: - 3 moles of CO - 3 moles of H2O - Let \( x \) be the number of moles of CO2 added. ### Step 3: Set up the equilibrium concentrations At equilibrium, we want the concentration of CO to be 2 M in a 1 L container. Therefore, the number of moles of CO at equilibrium will be: \[ \text{Moles of CO at equilibrium} = 2 \text{ moles} \] Since we started with 3 moles of CO, the change in moles of CO will be: \[ \text{Change in moles of CO} = 3 - 2 = 1 \text{ mole} \] ### Step 4: Determine changes in other species From the stoichiometry of the reaction: - 1 mole of CO is consumed, which means 1 mole of CO2 is produced. - Therefore, the moles of CO2 at equilibrium will be: \[ \text{Moles of CO}_2 = x + 1 \] ### Step 5: Write the expression for the equilibrium constant (Kc) The equilibrium constant \( K_c \) is given by: \[ K_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} \] Substituting the equilibrium concentrations: - Concentration of CO at equilibrium = 2 M - Concentration of H2O at equilibrium = \( 3 - 1 = 2 \) M (since 1 mole of H2O is consumed) - Concentration of CO2 at equilibrium = \( x + 1 \) M - Concentration of H2 at equilibrium = 1 M (since 1 mole of H2 is produced) Thus, we have: \[ K_c = \frac{(x + 1)(1)}{(2)(2)} = 5 \] ### Step 6: Solve for x From the equation: \[ \frac{x + 1}{4} = 5 \] Multiplying both sides by 4: \[ x + 1 = 20 \] Subtracting 1 from both sides: \[ x = 19 \] ### Conclusion The number of moles of CO2 that must be added is **19 moles**. ---
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL EQUILIBRIUM

    NARENDRA AWASTHI|Exercise Level 1 (Q.93 To Q.122)|1 Videos
  • CHEMICAL EQUILIBRIUM

    NARENDRA AWASTHI|Exercise Level 2|1 Videos
  • ATOMIC STUCTURE

    NARENDRA AWASTHI|Exercise Exercise|273 Videos
  • DILUTE SOLUTION

    NARENDRA AWASTHI|Exercise Level 3 - Match The Column|1 Videos

Similar Questions

Explore conceptually related problems

Write the equilibrium constant of the reaction C(s)+H_(2)O(g)hArrCO(g)+H_(2)(g)

The equilibrium constant K_(c) for the reaction, 2NaHCO_(3)(s)hArrNa_(2)CO_(3)(s)+CO_(2)(g)+H_(2)O(g)

At a certain temperature , the equilibrium constant (K_(c)) is 4//9 for the reaction : CO(g)+H_(2)O(g) hArr CO_(2)(g)+H_(2)(g) If we take 10 mole of each of the four gases in a one - litre container, what would be the equilibrium mole percent of H_(2)(g) ?

For the reaction, CO(g)+H_(2)O(g)hArrCO_(2)(g)+H_(2)(g) , at a given temperature, the equilibrium amount of CO_(2)(g) can be increased by

The equilibrium constant for the reaction CO(g) + H_(2)O(g)hArrCO_(2)(g) + H_(2)(g) is 3 at 500 K. In a 2 litre vessel 60 gm of water gas [equimolar mixture of CO(g) and H_(2) (g)] and 90 gm of steam is initially taken. What is the equilibrium concentration of H_(2)(g) at equilibrium (mole/L) ?

For the reaction CO(g) +H_(2)O(g) hArr CO_(2)(g)+H_(2)(g) at a given temperature the equilibrium amount of CO_(2) (g) can be increased by

For the reaction CO(g)+H_(2)O(g) hArr CO_(2)(g)+H_(2)(g) at a given temperature, the equilibrium amount of CO_(2)(g) can be increased by