For the equilibrium system CO (g) + 3 H2 (g) ⇌ H2O (g) + CH4 (g),...

For the equilibrium system CO (g) + 3 H2 (g) ⇌ H2O (g) + CH4 (g), ΔH = -206 kJ/mol, state whether the yield of CH4 will increase, decrease or remain the same for the following changes:

Adding some H2 gas to the system

Removing H2O from the system

Increasing the volume of the system

Decreasing the temperature on the system

Adding a catalyst to the system

Solutions

Expert Solution

Adding reactant will shift the reaction in the direction of product as per Le chatelier Principle

So, Equilibrium moves to product side

Answer: Increase

Removing product will shift the reaction in the direction of product as per Le chatelier Principle

So, Equilibrium moves to product side

Answer: Increase

Increasing volume will decrease the pressure which in turn shift the reaction in a direction which have greater gaseous molecules as per Le chatelier Principle

Here reactant has more gaseous molecule

So equilibrium will move to left

So, Equilibrium moves to reactant side

Answer: decrease

Decreasing Temperature will shift the reaction in a direction which release heat as per Le chatelier Principle

Forward reaction is exothermic in nature

hence, forward reaction will be favoured

So, Equilibrium moves to product side

Answer: Increase

Catalyst doesn't affect equilibrium

So, No effect on equilibrium

Answer: remains the same

queen_honey_blossom answered 1 year ago

Next > < Previous

Related Solutions

The system CO2(g) + H2(g) H2O(g) + CO(g) is at equilibrium at some temperature. At equilibrium,...

The system CO2(g) + H2(g) H2O(g) + CO(g) is at equilibrium at some temperature. At equilibrium, a 4.00L vessel contains 1.00 mole CO2, 1.00 mole H2, 2.40 moles H2O, and 2.40 moles CO. How many moles of CO2 must be added to this system to bring the equilibrium CO concentration to 0.669 mol/L? A. 0.498 moles B. 0.993 moles C. 0.429 moles D. 0.069 moles

Consider the reaction CO(g) + 3 H2 (g) ⇌ CH4 (g) + H2O(g) Kc = 7.7×10–23...

Consider the reaction CO(g) + 3 H2 (g) ⇌ CH4 (g) + H2O(g) Kc = 7.7×10–23 at 25 °C. 1.50 mol CH4 and 2.50 mol H2O are added to an empty 2.00-L container at 25 °C and allowed to reach equilibrium. What are the equilibrium amounts (in moles) of each species? The answer is: 1.5 mol CO, 4.5 mol H2 , 1.0 mol H2O, and 2.6×10–21 mol CH4. Please include written work and explanation :-) Thank you!

Consider the reaction CO(g) + 3 H2 (g) ⇌ CH4 (g) + H2O(g) Kc = 7.7×10–23...

Calculate the equilibrium composition for the reaction Cgraphite + H2O(g) _<->CO(g) + H2(g) at T =...

Calculate the equilibrium composition for the reaction Cgraphite + H2O(g) _<->CO(g) + H2(g) at T = 298 K, 500 K, 1000 K, and 2000 K and a pressure of 1 bar. For the initial number of moles, use 1 mole of graphite, 3 moles of H2O(g), and no CO or H2.

23.Consider the following equilibrium process at 686°C: CO2(g) + H2(g) ⇌ CO(g) + H2O(g) The equilibrium...

23.Consider the following equilibrium process at 686°C: CO2(g) + H2(g) ⇌ CO(g) + H2O(g) The equilibrium concentrations of the reacting species are [CO] = 0.0580 M, [H2] = 0.0430 M, [CO2] = 0.0900 M, and [H2O] = 0.0420 M. (a) Calculate Kc for the reaction at 686°C. ________ (b) If we add CO2 to increase its concentration to 0.460 mol / L, what will the concentrations of all the gases be when equilibrium is reestablished? CO2: M H2: M CO:...

PS8.2. The equilibrium constant, KP, for the reaction CO2(g) + H2(g) H2O(g) + CO(g) is 0.138....

PS8.2. The equilibrium constant, KP, for the reaction CO2(g) + H2(g) H2O(g) + CO(g) is 0.138. Calculate the partial pressure of all species at equilibrium for each of the following original mixtures: a) 1.36 atm of CO2 and 1.36 atm of H2. b) 0.87 atm of CO2, 0.87 atm of H2 and 0.87 atm of H2O(g). c) 0.64 atm of H2O and 0.64 atm of CO.

Consider the equilibrium CH4(g) + H2O(g) <--> CO(g) + 3H2(g), where delta h= 206 KJ. Which...

Consider the equilibrium CH4(g) + H2O(g) <--> CO(g) + 3H2(g), where delta h= 206 KJ. Which of the following disturbances will NOT cause the system to shift to the right to reestablish equilibrium? A.) The partial pressure of CH4 increases. B.) The partial pressure of CO decreases. C.) The volume decreases. D.) The temperature increases. E.) All of these will cause the system to shift to the right.

Consider the following equilibrium: CO2(g) + H2(g)---CO(g) + H2O(g); Kc = 1.6 at 1260 K Suppose...

Consider the following equilibrium: CO2(g) + H2(g)---CO(g) + H2O(g); Kc = 1.6 at 1260 K Suppose 0.038 mol CO2 and 0.022 mol H2 are placed in a 1.50-L vessel at 1260 K. What is the equilibrium partial pressure of CO(g)? The answer is either 9.9 atm or 1.1 atm. But my question is WHY IS ONE CHOICE INCORRECT since I get both in my calculations.

3. C (s) + H2O (g) → CO (g) + H2 (g) Le Châtelier's principle predicts...

3. C (s) + H2O (g) → CO (g) + H2 (g) Le Châtelier's principle predicts that increasing the volume of the reaction vessel will ________.

Consider the reaction: CO(g) + H2O(g)<------> CO(g) + H2(g) and K = 0.118 at 4000 K....

Consider the reaction: CO(g) + H2O(g)<------> CO(g) + H2(g) and K = 0.118 at 4000 K. A reaction mixture initially contains a CO partial pressure of 1344 mbar and a H2O partial pressure of 1766 mbar at 4000K. Calculate the equilibrium partial pressures of each of the products.

Subjects