At a given temperature, K = 1.3x10^-2 for the reaction: N2(g) + 3H2(g) ⇌ 2NH3(g)

Calculate values of K for the following reactions at this temperature.
a. 1/2N2 + 3/2H2(g) ⇌ NH3(g)
b. 2NH3(g) ⇌ N2(g) + 3H2(g)
c. NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)
d. 2N2(g) + 6H2(g) ⇌ 4NH3(g)

a) 0.11

b)76.9

c) 8.8

d) 1.7*10^-4

Explanation:

Step 1: Data given

K = 1.3 * 10^-2 for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g)

Step 2: Formula of K

aA(g) + bB(g) ⇌ cC(g) + dD(g)

K = [C]^c *[D]^d  / [A]^a * [B]^b

K = 1.3 * 10^-2 = [NH3]² / [H2]³*[N2]

Step 3:

a) 1/2N2 + 3/2H2(g) ⇌ NH3(g)

N2(g) + 3H2(g) ⇌ 2NH3

1/2N2 + 3/2H2(g) ⇌ NH3(g)    =>K' =  

K' = = 0.11

b. 2NH3(g) ⇌ N2(g) + 3H2(g)

N2(g) + 3H2(g) ⇌ 2NH3

2NH3(g) ⇌ N2(g) + 3H2(g)    =>K' = 1/K

K' = 1/(1.3*10^-2) = 76.9

c. NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)

N2(g) + 3H2(g) ⇌ 2NH3

NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)    

=>K' =

K' =

K' = 8.8

d. 2N2(g) + 6H2(g) ⇌ 4NH3(g)

N2(g) + 3H2(g) ⇌ 2NH3

2N2(g) + 6H2(g) ⇌ 4NH3(g)

K' = K²

K' = (1.3*10^-2)²

K' = 1.7 *10 ^-4

i think it's c. i'm not sure but i think so.

because of all the things that happen around the us, many people get into arguments which sadly could get physical, but we should actually make a change for that.