A 1.0 L vessel at 300 K contains 0.368 g of N2O4(g) and 0.368 g of NO2(g) (molar masses 92 and 46 g/mol). For N2O4(g)

Question

A 1.0 L vessel at 300 K contains 0.368 g of N2O4(g) and 0.368 g of NO2(g) (molar masses 92 and 46 g/mol). For N2O4(g) <=> 2NO2(g), K_p = 0.15 atm at 300 K. Take R = 0.0821 L atm mol^-1 K^-1. Which conclusion is correct?

Accepted Answer

The reaction proceeds backward, since Q_c = 0.016 mol L^-1 exceeds K_c = 0.0061 mol L^-1

Answer

No conclusion is possible without the total pressure

Answer

The reaction proceeds forward, since Q_c = 0.016 is smaller than K_p = 0.15

Answer

The mixture is at equilibrium, because the masses of the two gases are equal

A 1.0 L vessel at 300 K contains 0.368 g of $N X_{2} O X_{4} (g)$ and 0.368 g of $NO X_{2} (g)$ (molar masses 92 and 46 g/mol). For $N X_{2} O X_{4} (g) 2 NO X_{2} (g)$ , $K_{p} = 0.15$ atm at 300 K. Take $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ . Which conclusion is correct?

Answer: D

More questions on Relation between Kp and Kc, units

A 1.0 L vessel at 300 K contains 0.368 g of NX2​OX4​(g) and 0.368 g of NOX2​(g) (molar masses 92 and 46 g/mol). For NX2​OX4​(g)​2NOX2​(g), Kp​=0.15 atm at 300 K. Take R=0.0821 L atm mol−1 K−1. Which conclusion is correct?

Answer: D

More questions on Relation between Kp and Kc, units

A 1.0 L vessel at 300 K contains 0.368 g of $N X_{2} O X_{4} (g)$ and 0.368 g of $NO X_{2} (g)$ (molar masses 92 and 46 g/mol). For $N X_{2} O X_{4} (g) 2 NO X_{2} (g)$ , $K_{p} = 0.15$ atm at 300 K. Take $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ . Which conclusion is correct?