# Problem: The oxidation of nitrogen monoxide is favored at 457 K:           2NO(g) + O 2(g) ⇌ 2NO2(g)           Kp = 1.3×104(a) Calculate Kc at 457 K.

###### FREE Expert Solution

2 NO(g) + O2(g) ⇌ 2NO2(g)           Kp = 1.3×104

$\overline{){{\mathbf{K}}}_{{\mathbf{p}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{c}}}{\left(\mathbf{RT}\right)}^{\mathbf{∆}\mathbf{n}}}$

R = gas constant = 0.08206 (Latm)/(molK)
T = temperature, K
Δn = moles of gas products – moles of gas reactants

Calculate K­c:

Given:                        Kp = 1.3×104

T = 457 K

Reaction:

2 NO(g) + O2(g) ⇌ 2 NO2(g)

reactants = 2 mol NO(g) + 1 mol O2(g)
reactants = 3 moles gas

products = 2 NO2(g)
products = 2 moles gas

Δn = 2 mol – 3 mol
Δn = –1 mol

92% (197 ratings) ###### Problem Details

The oxidation of nitrogen monoxide is favored at 457 K:
2NO(g) + O 2(g) ⇌ 2NO2(g)           Kp = 1.3×104
(a) Calculate Kc at 457 K.

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