The equilibrium constant for reaction (1) below is 276. Under the same conditions, what is the equilibrium constant of reaction (2) ? (1) ½X2(g) + ½Y2(g) 
XY(g)
(2) 2XY(g) 
X2(g) + Y2(g)
A) 6.02 × 10-2
B) 7.25 × 10-3
C) 3.62 × 10-3
D) 1.31 × 10-5
E) None of these choices are correct.
Correct Answer:
Verified
Q18: Write the mass-action expression, Q19: Consider the equilibrium reaction shown below. B2(g) Q20: Write the mass-action expression, Q21: Consider the following two equilibria and their Q22: Write the mass-action expression, Q24: The equilibrium constant, Kc, for the decomposition Q25: The equilibrium constant, Kp, for the reaction Q26: What is the mass-action expression, Q27: The reaction of nitrogen with oxygen to Q28: Write the mass-action expression, Unlock this Answer For Free Now! View this answer and more for free by performing one of the following actions Scan the QR code to install the App and get 2 free unlocks Unlock quizzes for free by uploading documents
