For $N X_{2} O X_{4} (g) 2 NO X_{2} (g)$ at 298 K, $K_{c} = 5.8 \times 1 0^{- 3} mol L^{- 1}$ and $K_{p} = 0.142$ atm (use $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ ). Let $K_{c}^{'}$ and $K_{p}^{'}$ refer to $\frac{1}{2} N X_{2} O X_{4} (g) NO X_{2} (g)$ . Which statements are correct?

Answer: B, C, D

Halving an equation takes the square root of

K

K_{c}^{'} = K_{c} = 0.0058 = 0.076

, not

K_{c} /2

. Reversing inverts

K

1/ K_{p} = 1/0.142 = 7.04 at m^{- 1}

. For the halved reaction

Δ n = 1 - \frac{1}{2} = \frac{1}{2}

, so

K_{p}^{'} = K_{c}^{'} (RT)^{1/2}

. Statements 2, 3 and 4 are correct.