The equlibrium constant K, for the reaction $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ is $1.64 \times 10 - 4 a t m$ at $400^{\circ} C$ . What will be the equilibrium constant at $500^{\circ} C$ , if heat of reaction in this temperature range is - 105185.8 Joules.

The equlibrium constant K, for the reaction $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ is $1.64 \times 10_{4} a t m$ at $400^{\circ} C$ . What will be the equilibrium constant at $500^{\circ} C$ , if heat of reaction in this temperature range is - 105185.8 Joules.

Q. If

K_{1}

is the equilibrium constant for the reaction

2 N H_{3} ⇌ N_{2} + 3 H_{2}

, then equilibrium constant for the reaction,

\frac{1}{2} N_{2} + \frac{3}{2} H_{2} ⇌ N H_{3}

will be

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What will be the equilibrium constant at 127oC. If equilibrium constant at 27oC is 4 for reaction N2+3H2⇌2NH3;ΔH=−46.06kJ.

The correct option is C 4×10−2By using the Vanthoff's equation log K2K1 = ΔH2.303R x T2−T1T1T2hereK1 = 4 ΔH = -46.06 kJ T2 = 400 K T1 = 300 K R is gas constant (0.008314)by solving that we will get equilibrium constant K2 as 0.04So, the correct option is 4×10−2.

What will be the equilibrium constant at $127^{o}$ C. If equilibrium constant at $27^{o}$ C is $4$ for reaction $N_{2} + 3 H_{2} ⇌ 2 N H_{3}; Δ H = - 46.06$ kJ.