# Chapter 12 - Chemical Equilibrium - Exercise 12.5 - Relating K(c) and K(p) - Page 535: b

$$K_p = 1.4 \times 10^{80} \space atm^{-1}$$

#### Work Step by Step

1. $T(K) = 25 + 273 = 298$ 2. $\Delta n = 2 - 2 - 1 = -1$ 3. $$K_p = K_c \times RT^{\Delta n} = (3.5 \times 10^{81}) \times (0.0821 \space L \space atm \space mol^{-1} \space K^{-1} \times 298 \space K)^{-1}$$ $$K_p = 1.4 \times 10^{80} \space atm^{-1}$$

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