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The reaction 2 NO(g) ⇌ N2(g) + O2(g) has a Kc value of 2400 at 2000 K. If 0.850 M each of N2 and O2 are initially present in a 3.00-L vessel, calculate the equilibrium concentrations of NO, N2, and O2.​

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Magansinoviz

Answer:

O2(g)=  8.41x10^-1

N2(g)=  8.41x10^-1

NO(g)= 1.72x10^-2

Explanation:

2 NO(g) ⇌ N2(g) + O2(g)  Kc = 2400

                        2 NO(g) ⇌ N2(g) + O2(g)

Initial                      0         0.850     0.850

Change             +2x           -x            -x

Equilibrium        2x        0.850-x     0.850-x

Kc= [N2][O2]/[NO]^2

2400= (0.850-x )^2/ 4x^2

x= 0.00858763

O2(g)=  0.850-0.00858763 = 8.41x10^-1

N2(g)=  0.850-0.00858763 = 8.41x10^-1

NO(g)= 2(0.00858763) = 1.72x10^-2

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