When 5.1g of solid NH₄HS is introduced into a two litre evacuated flask at 27°C, 20% of the solid decomposes into gaseous ammonia and hydrogen sulphide. The K_p for the reaction at 27°C is x × 10^-2. The value of x is_______. (Integer answer)

[Given R = 0.082 L atm K^-1 mol^-1]

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alok kumar singh | Contributor-Level 10

3 months ago

Moles of NH₄HS initially taken = $\frac{5.1}{51} = 0.1 m o l$

$N H_{4} H S (S) + N H_{3} (g) + H_{2} S (g)$

t = $\infty$ 0 0.1 mol 0 0

t = $\infty$ 0.1 (1 - 0.2) 0.1 × 0.2 0.1 × 0.2

$P_{N H_{3}} = \frac{n R T}{V} = \frac{0.1 \times 0.2 \times 0.082 \times 300}{2} = 0.246 a t m = P_{H_{2} S}$

$K_{P} = P_{N H_{3}} \times P_{H_{2} S} = {(0.246)}^{2} = 0.060516 = 6.05 \times 1 0^{- 2}$

x = 6 (Nearest integer).

Moles of NH4HS initially taken = <math> <mrow> <mfrac> <mrow> <mn>5</mn> <mo>.</mo> <mn>1</mn> </mrow> <mrow> <mn>5</mn> <mn>1</mn> </mrow> </mfrac> <mo>=</mo> <mn>0</mn> <mo>.</mo> <mn>1</mn> <mi>m</mi> <mi>o</mi> <mi>l</mi> </mrow> </math>   <math> <mrow> <mi>N</mi> <msub> <mrow> <mi>H</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msub> <mi>H</mi> <mi>S</mi> <mrow> <mo>(</mo> <mrow> <mi>S</mi> </mrow> <mo>)</mo> </mrow> <mo>+</mo> <mi>N</mi> <msub> <mrow> <mi>H</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msub> <mrow> <mo>(</mo> <mrow> <mi>g</mi> </mrow> <mo>)</mo> </mrow> <mo>+</mo> <msub> <mrow> <mi>H</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mi>S</mi> <mrow> <mo>(</mo> <mrow> <mi>g</mi> </mrow> <mo>)</mo> </mrow> </mrow> </math>                             t =  <math> <mrow> <mo>∞</mo> </mrow> </math> 0       0.1 mol               0            0t =   <math> <mrow> <mo>∞</mo> </mrow> </math>  0.1 (1 - 0.2)   0.1 × 0.2   0.1 × 0.2 <math> <mrow> <msub> <mrow> <mi>P</mi> </mrow> <mrow> <mi>N</mi> <msub> <mrow> <mi>H</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msub> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mi>n</mi> <mi>R</mi> <mi>T</mi> </mrow> <mrow> <mi>V</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>0</mn> <mo>.</mo> <mn>1</mn> <mo>×</mo> <mn>0</mn> <mo>.</mo> <mn>2</mn> <mo>×</mo> <mn>0</mn> <mo>.</mo> <mn>0</mn> <mn>8</mn> <mn>2</mn> <mo>×</mo> <mn>3</mn> <mn>0</mn> <mn>0</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>0</mn> <mo>.</mo> <mn>2</mn> <mn>4</mn> <mn>6</mn> <mi>a</mi> <mi>t</mi> <mi>m</mi> <mo>=</mo> <msub> <mrow> <mi>P</mi> </mrow> <mrow> <msub> <mrow> <mi>H</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mi>S</mi> </mrow> </msub> </mrow> </math> <math> <mrow> <msub> <mrow> <mi>K</mi> </mrow> <mrow> <mi>P</mi> </mrow> </msub> <mo>=</mo> <msub> <mrow> <mi>P</mi> </mrow> <mrow> <mi>N</mi> <msub> <mrow> <mi>H</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msub> </mrow> </msub> <mo>×</mo> <msub> <mrow> <mi>P</mi> </mrow> <mrow> <msub> <mrow> <mi>H</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mi>S</mi> </mrow> </msub> <mo>=</mo> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mn>0</mn> <mo>.</mo> <mn>2</mn> <mn>4</mn> <mn>6</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mn>0</mn> <mo>.</mo> <mn>0</mn> <mn>6</mn> <mn>0</mn> <mn>5</mn> <mn>1</mn> <mn>6</mn> <mo>=</mo> <mn>6</mn> <mo>.</mo> <mn>0</mn> <mn>5</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>2</mn> </mrow> </msup> </mrow> </math> x = 6 (Nearest integer).

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When 5.1g of solid NH₄HS is introduced into a two litre evacuated flask at 27°C, 20% of the solid decomposes into gaseous ammonia and hydrogen sulphide. The K_p for the reaction at 27°C is x × 10^-2. The value of x is_______. (Integer answer)

[Given R = 0.082 L atm K^-1 mol^-1]

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When 5.1g of solid NH4HS is introduced into a two litre evacuated flask at 27°C, 20% of the solid decomposes into gaseous ammonia and hydrogen sulphide. The Kp for the reaction at 27°C is x × 10-2. The value of x is_______. (Integer answer) [Given R = 0.082 L atm K-1 mol-1]

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When 5.1g of solid NH₄HS is introduced into a two litre evacuated flask at 27°C, 20% of the solid decomposes into gaseous ammonia and hydrogen sulphide. The K_p for the reaction at 27°C is x × 10^-2. The value of x is_______. (Integer answer)

[Given R = 0.082 L atm K^-1 mol^-1]