When 5.1g of solid $N H_{4} H S$ is introduced a two litre evacuated flask at 270C, 20% of the solid decomposes into gaseous ammonia and hydrogen sulphide. The Kp for the reaction at 27 degree C is x $\times 10^{- 2}$ , The value of x is .

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answer is 6.

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Detailed Solution

It is given to idenify x.

Explanation $P = \frac{n R T}{V} = \frac{0.1 \times 0.2 \times 0.082 \times 300}{2} = 0.246 K_{p} = P_{N H_{3}} \times P_{H_{2} S} = {(0.246)}^{2} = 0.060516 = 6.05 \times 10^{5} \approx 6$

Hence, the correct answer 6.

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