Given the following: i) n20(g) 1/2 02(g) 2 no(g) ii) n2(g) 02(g) 2 no(g) ke= 4.1 x 10-31 ke= 1.7 x 10-13 find the value of the equilibrium constant for the following equilibrium reaction: n2(g) 1/202(g) n20(g) a) 7.0 x 10-44 b) 4.2 x 1017 c) 2.4 x 10-18 d) 1.6 x 109 e) 2.6 x 10-22

Answer B) 4.2x10^17

Explanation:

To produce the reaction 3 using reaction 1 and 2 we need to invert the order of the first reaction the second in the same order, as it's shown:

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Due to the inversion of the first equation, the equilibrium constant of the new reaction is K1'=1/K1.=2.4x10^30

Finally, the new equilibrium constant K3 is the product of the previous constants:  

K3=K1'*K2=4.2x10^17

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"