At 30- C- the gaseous mixture for the equilibrium- N2g - O2g - 2NOg- contains 56 gm of N2- 128 gm of O2 and 120 gm of NO at one atmospheric pressure- The value of Kp at 30- C is-

A vessel of volume 4 litres contains a mixture of 8gms of O $_{2}$ , 14 gms of N $_{2}$ and 22 gms of CO $_{2}$ at 27 $^{0}$ C. The pressure exerted by the mixture is

Δ G = 77.77 k J {m o l}^{- 1}

1000 K

for the reaction

\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) = N O (g)

. Partial pressure of

N O

under equilibrium at

1000 K

for air at 1 atm pressure containing 80%

N_{2}

and 20%

O_{2}

by volume is

X \times 10^{- 5}

atm.

100 X

is__________.

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	$N_{2}$	$O_{2}$	$N O$
Mass (g)	56	128	120
Mol wt (g/mol)	28	32	30
Moles	2	4	4
Partial pressure (atm)	$\frac{2}{10} \times 1 = 0.2$ ,	$\frac{4}{10} \times 1 = 0.4$ ,	$\frac{4}{10} \times 1 = 0.4$

At 30∘C, the gaseous mixture for the equilibrium, N2(g)+O2(g)⇔2NO(g), contains 56 gm of N2, 128 gm of O2 and 120 gm of NO at one atmospheric pressure. The value of Kp at 30∘C is:

The correct option is A 2The following table lists the quantities of various species present:N2O2NOMass (g)56128120Mol wt (g/mol)283230Moles244Partial pressure (atm)210×1=0.2, 410×1=0.4, 410×1=0.4The expression for the equilibrium constant becomes, Kp=P2NOPN2PO2=(0.4)20.2×0.4=2.

At $30^{\circ} C$ , the gaseous mixture for the equilibrium, $N_{2} (g) + O_{2} (g) \Leftrightarrow 2 N O (g)$ , contains 56 gm of $N_{2}$ , 128 gm of $O_{2}$ and 120 gm of $N O$ at one atmospheric pressure. The value of $K_{p}$ at $30^{\circ} C$ is: