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5.1g NH4SH is introduced in 3.0L evacuated flask at 327degree celcius ,30percent of the solid NH4SH decomposed to NH3 and H2S as gases. The Kp of the reaction at 327degree celcius is ?(R=0.082Latm mol-1K-1,molar mass ofS=32gmol-1,molar mass of N =14gmol-1).((guys I need the answer with full explanation please answer me. I need your help))?
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5.1g NH4SH is introduced in 3.0L evacuated flask at 327degree celcius ...
**Given Information:**
- Mass of NH4SH = 5.1g
- Volume of flask = 3.0L
- Temperature = 327°C
- Percentage of NH4SH decomposed = 30%
- R = 0.082 L atm mol^(-1) K^(-1)
- Molar mass of S = 32 g/mol
- Molar mass of N = 14 g/mol
**Calculating the moles of NH4SH:**
Molar mass of NH4SH = (14 g/mol * 2) + (1 g/mol * 4) + (32 g/mol) = 32 g/mol
Moles of NH4SH = mass / molar mass = 5.1 g / 32 g/mol ≈ 0.159 mol
**Calculating the moles of NH3 and H2S:**
Since 30% of NH4SH decomposed, only 70% will remain.
Moles of NH4SH remaining = 0.159 mol * 0.70 = 0.111 mol
Moles of NH3 and H2S produced = 0.159 mol - 0.111 mol = 0.048 mol
**Calculating the partial pressure of NH3 and H2S:**
Volume of flask = 3.0 L
Partial pressure = moles * R * temperature / volume
Partial pressure of NH3 = (0.048 mol * 0.082 L atm mol^(-1) K^(-1) * (327 + 273) K) / 3.0 L = 0.623 atm
Partial pressure of H2S = (0.048 mol * 0.082 L atm mol^(-1) K^(-1) * (327 + 273) K) / 3.0 L = 0.623 atm
**Calculating the equilibrium constant (Kp):**
The equilibrium constant (Kp) can be determined by using the partial pressures of the reactants and products.
The reaction can be represented as:
NH4SH(s) ⇌ NH3(g) + H2S(g)
Kp = (P(NH3) * P(H2S)) / P(NH4SH)
Substituting the values:
Kp = (0.623 atm * 0.623 atm) / 0.623 atm = 0.623
Therefore, the equilibrium constant (Kp) of the reaction at 327°C is approximately 0.623.
- Mass of NH4SH = 5.1g
- Volume of flask = 3.0L
- Temperature = 327°C
- Percentage of NH4SH decomposed = 30%
- R = 0.082 L atm mol^(-1) K^(-1)
- Molar mass of S = 32 g/mol
- Molar mass of N = 14 g/mol
**Calculating the moles of NH4SH:**
Molar mass of NH4SH = (14 g/mol * 2) + (1 g/mol * 4) + (32 g/mol) = 32 g/mol
Moles of NH4SH = mass / molar mass = 5.1 g / 32 g/mol ≈ 0.159 mol
**Calculating the moles of NH3 and H2S:**
Since 30% of NH4SH decomposed, only 70% will remain.
Moles of NH4SH remaining = 0.159 mol * 0.70 = 0.111 mol
Moles of NH3 and H2S produced = 0.159 mol - 0.111 mol = 0.048 mol
**Calculating the partial pressure of NH3 and H2S:**
Volume of flask = 3.0 L
Partial pressure = moles * R * temperature / volume
Partial pressure of NH3 = (0.048 mol * 0.082 L atm mol^(-1) K^(-1) * (327 + 273) K) / 3.0 L = 0.623 atm
Partial pressure of H2S = (0.048 mol * 0.082 L atm mol^(-1) K^(-1) * (327 + 273) K) / 3.0 L = 0.623 atm
**Calculating the equilibrium constant (Kp):**
The equilibrium constant (Kp) can be determined by using the partial pressures of the reactants and products.
The reaction can be represented as:
NH4SH(s) ⇌ NH3(g) + H2S(g)
Kp = (P(NH3) * P(H2S)) / P(NH4SH)
Substituting the values:
Kp = (0.623 atm * 0.623 atm) / 0.623 atm = 0.623
Therefore, the equilibrium constant (Kp) of the reaction at 327°C is approximately 0.623.
Community Answer
5.1g NH4SH is introduced in 3.0L evacuated flask at 327degree celcius ...
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5.1g NH4SH is introduced in 3.0L evacuated flask at 327degree celcius ,30percent of the solid NH4SH decomposed to NH3 and H2S as gases. The Kp of the reaction at 327degree celcius is ?(R=0.082Latm mol-1K-1,molar mass ofS=32gmol-1,molar mass of N =14gmol-1).((guys I need the answer with full explanation please answer me. I need your help))?
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5.1g NH4SH is introduced in 3.0L evacuated flask at 327degree celcius ,30percent of the solid NH4SH decomposed to NH3 and H2S as gases. The Kp of the reaction at 327degree celcius is ?(R=0.082Latm mol-1K-1,molar mass ofS=32gmol-1,molar mass of N =14gmol-1).((guys I need the answer with full explanation please answer me. I need your help))? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 5.1g NH4SH is introduced in 3.0L evacuated flask at 327degree celcius ,30percent of the solid NH4SH decomposed to NH3 and H2S as gases. The Kp of the reaction at 327degree celcius is ?(R=0.082Latm mol-1K-1,molar mass ofS=32gmol-1,molar mass of N =14gmol-1).((guys I need the answer with full explanation please answer me. I need your help))? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5.1g NH4SH is introduced in 3.0L evacuated flask at 327degree celcius ,30percent of the solid NH4SH decomposed to NH3 and H2S as gases. The Kp of the reaction at 327degree celcius is ?(R=0.082Latm mol-1K-1,molar mass ofS=32gmol-1,molar mass of N =14gmol-1).((guys I need the answer with full explanation please answer me. I need your help))?.
5.1g NH4SH is introduced in 3.0L evacuated flask at 327degree celcius ,30percent of the solid NH4SH decomposed to NH3 and H2S as gases. The Kp of the reaction at 327degree celcius is ?(R=0.082Latm mol-1K-1,molar mass ofS=32gmol-1,molar mass of N =14gmol-1).((guys I need the answer with full explanation please answer me. I need your help))? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 5.1g NH4SH is introduced in 3.0L evacuated flask at 327degree celcius ,30percent of the solid NH4SH decomposed to NH3 and H2S as gases. The Kp of the reaction at 327degree celcius is ?(R=0.082Latm mol-1K-1,molar mass ofS=32gmol-1,molar mass of N =14gmol-1).((guys I need the answer with full explanation please answer me. I need your help))? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5.1g NH4SH is introduced in 3.0L evacuated flask at 327degree celcius ,30percent of the solid NH4SH decomposed to NH3 and H2S as gases. The Kp of the reaction at 327degree celcius is ?(R=0.082Latm mol-1K-1,molar mass ofS=32gmol-1,molar mass of N =14gmol-1).((guys I need the answer with full explanation please answer me. I need your help))?.
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