Asked by ami
                Consider the equilibrium of water with gaseous hydrogen and oxygen: 
 
Give the value of for this reaction, if it is written as:
 
Choose one answer.
 
a. -7.8 x 10^-16
 
b. -3.9 x 10^-16
 
c. 1.3 x 10^-15
 
d. 6.4 x 10^-14
 
e. 3.6 x 10^2
            
            
        Give the value of for this reaction, if it is written as:
Choose one answer.
a. -7.8 x 10^-16
b. -3.9 x 10^-16
c. 1.3 x 10^-15
d. 6.4 x 10^-14
e. 3.6 x 10^2
Answers
                    Answered by
            DrBob222
            
    give the value of WHAT for this reaction, ......
                    Answered by
            ami
            
    Consider the equilibrium of water with gaseous hydrogen and oxygen: 
2H2O(g)=2H2(g) + O2(g) Kp= 7.8x10^-16
Give the value of for this reaction, if it is written as:
1/2O2(g) + H2(g) =H2O(g)
 
Choose one answer.
 
a. -7.8 x 10^16
 
b. -3.9 x 10^16
 
c. 1.3 x 10^15
 
d. 6.4 x 10^14
 
e. 3.6 x 10^7
    
2H2O(g)=2H2(g) + O2(g) Kp= 7.8x10^-16
Give the value of for this reaction, if it is written as:
1/2O2(g) + H2(g) =H2O(g)
Choose one answer.
a. -7.8 x 10^16
b. -3.9 x 10^16
c. 1.3 x 10^15
d. 6.4 x 10^14
e. 3.6 x 10^7
                    Answered by
            DrBob222
            
    You still didn't say "of what" but I assume it is Kp after you filled in th remainder of the missing information.
First you turned the reaction around. Let's designate the original Kp as Kp1. That makes the new Kp (which we will call Kp2) = 1/Kp1. THEN you divided all of the coefficients by 2. That makes Kp3 = sqrt(Kp2). Therefore, you take the reciprocal of the original Kp, then take the square root of that.
    
First you turned the reaction around. Let's designate the original Kp as Kp1. That makes the new Kp (which we will call Kp2) = 1/Kp1. THEN you divided all of the coefficients by 2. That makes Kp3 = sqrt(Kp2). Therefore, you take the reciprocal of the original Kp, then take the square root of that.
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