Consider the following reaction:
C(s) + H2O (g) ⇌ CO (g) + H2 (g) (Kp = 0.45 at 900 K)
What is the equilibrium partial pressure of H2O when the initial partial pressure of water is 1.00 atm?
a) 0.43 atm
b) 0.58 atm
c) 0.22 atm
d) 0.52 atm
e) 1.16 atm
Be sure to read and follow the directions that are on the Post a New Question screen:
Homework Posting Tips
~ Please show your work. Tutors will not do your homework for you. Please show your work for any question that you are posting.
~ Please proofread your question. Does your question use proper grammar, capitalization and spelling?
~ Please be patient. All tutors are volunteers, and sometimes a tutor may not be immediately available. Please be patient while waiting for a response to your question.
.......C(s) + H2O(g) ⇌ CO(g) + H2(g)
I.............1.00......0.......0
C..............-x.......x.......x
E............1-x........x.......x
Substitute the E line into Kp expressin and solve for x = pH2 = pCO
pH2O = 1-x. The correct answer is D.
Thank you Dr.Bob222 for clarification.
I am reviewing all the information for Gen Chem 2 for a big exam next week; and as you know the information is quite vast and it takes a bit to recall how to solve the problems correctly. I do not seek for mere answers to these problems, nor for someone to do my homework (I do have the key to all of them), rather I am interested in reviewing, recalling and understanding the concepts. Again thank you Dr. Bob for your help.
Dear Writeacher,
Thank you for the attention.
However, following up on what Writeacher wrote, it is a huge help to us if you post what you think is the way to fo, post the process, explain what you don't understand, etc. That way we can focus on helping with the specific part of the problem that is causing concern rather than typing in a bunch of stuff you already know how to do. It's a much more efficient process if you start by showing what you already know.
True. Will do so
To find the equilibrium partial pressure of H2O, we need to use the given equilibrium constant (Kp) and the initial partial pressure of water. Here's how you can do it step by step:
Step 1: Write the equilibrium expression:
Kp = (PH2 * PCO) / (PH2O * PC)
Step 2: Plug in the given values:
Kp = 0.45 (given)
PH2O = 1.00 atm (given)
Step 3: Set up the equilibrium expression with the unknown variable:
0.45 = (PH2 * PCO) / (1.00 * PC)
Step 4: Rearrange the equation and solve for PH2:
PH2 = (0.45 * PC) / PCO
Step 5: Notice that the reaction stoichiometry tells us that for every 1 mole of CO formed, 1 mole of H2O reacts. Therefore, the initial partial pressure of CO (PCO) is equal to the initial partial pressure of H2O (PH2O).
Step 6: Substitute PCO with PH2O in the equation from Step 4:
PH2 = (0.45 * PH2O) / PH2O
Step 7: Simplify the equation:
PH2 = 0.45
Therefore, the equilibrium partial pressure of H2O is 0.45 atm.
None of the given answer choices match this value, so it seems there may be an error in the problem or the answer options provided. I recommend double-checking the given information or referring to the original source for accurate answer options.