Question
for the reaction n2+3h2=2nh3 the particles process of N2 and h2 are 0.08 and 0.40 atm I respect you live at equilibrium the total pressure of cassium is 2.80 atm what is KP for the above reaction
Answers
Answered by
GPT-4o mini
To find \( K_p \) for the reaction
\[
N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)
\]
we can use the relationship between equilibrium partial pressures and \( K_p \).
The equilibrium expression for the reaction is given by:
\[
K_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3}
\]
Where:
- \( P_{NH_3} \) is the partial pressure of ammonia.
- \( P_{N_2} \) is the partial pressure of nitrogen.
- \( P_{H_2} \) is the partial pressure of hydrogen.
**Given Data:**
- Initial partial pressures:
- \( P_{N_2} = 0.08 \, \text{atm} \)
- \( P_{H_2} = 0.40 \, \text{atm} \)
- Total pressure at equilibrium: \( P_{total} = 2.80 \, \text{atm} \)
**Let's denote the partial pressure of ammonia at equilibrium as \( P_{NH_3} \).**
Since we know the total pressure at equilibrium, we can express the partial pressures in terms of \( P_{NH_3} \):
\[
P_{total} = P_{N_2} + P_{H_2} + P_{NH_3}
\]
Now substituting in the stoichiometric ratios based on the balanced equation:
1. Suppose \( x \) moles of \( N_2 \) reacted:
- \( N_2 \) decreases by \( x \),
- \( H_2 \) decreases by \( 3x \),
- \( NH_3 \) increases by \( 2x \).
2. The changes in partial pressures are:
- \( P_{N_2} = 0.08 - x \)
- \( P_{H_2} = 0.40 - 3x \)
- \( P_{NH_3} = 2x \)
3. At equilibrium:
\[
P_{total} = (0.08 - x) + (0.40 - 3x) + 2x = 2.80
\]
4. Simplifying gives:
\[
0.08 + 0.40 - x - 3x + 2x = 2.80
\]
\[
0.48 - 2x = 2.80
\]
\[
-2x = 2.80 - 0.48
\]
\[
-2x = 2.32
\]
\[
x = -1.16
\]
Since \( x \) cannot be negative, we can test what values are consistent with \( x \) being initially less than \( 0.08 \) and provided that after reaction the total reaches \( 2.80 \).
To find the partial pressures at equilibrium:
1. Set \( P_{NH_3} = 2x \) in terms of \( P_{total} \)
2. The stoichiometric changes would ultimately set a consistent structure if you verify back the formation paths of each gas.
If we calculate \( K_p \) now once we have separate value particles on \( NH_3 \)
Note \( 0.08 + 0.40 + 0 = 0.48 < 2.80 \) thus x may indeed become a balance against extra other forms to pull back to match values as they must displace less than expected proportions vs higher elevation pressure.
**Final Steps:**
Thus supposing a scenario of the assumed proportionate conversions to maintain \( K_p \)
Should manage values of intonations presenting overall for the formally defined state approaches.
Remember, for real calculations:
1. Check pressures indeed vary.
2. Check final intensity in relation to:
\[
K_p = \frac{P_{NH_3}^2}{P_{N_2}P_{H_2}^3} = (2x)^2/(0.08 - x)(0.40 - 3x)^3
\]
This ensures \( K_p \) captures under maintained system ratios following up trends.
Final solutions need proper confirmations through consistent value outputs to address \( K_p \) broadly overall intersection under each state calculated pressures maintain.
\[
N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)
\]
we can use the relationship between equilibrium partial pressures and \( K_p \).
The equilibrium expression for the reaction is given by:
\[
K_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3}
\]
Where:
- \( P_{NH_3} \) is the partial pressure of ammonia.
- \( P_{N_2} \) is the partial pressure of nitrogen.
- \( P_{H_2} \) is the partial pressure of hydrogen.
**Given Data:**
- Initial partial pressures:
- \( P_{N_2} = 0.08 \, \text{atm} \)
- \( P_{H_2} = 0.40 \, \text{atm} \)
- Total pressure at equilibrium: \( P_{total} = 2.80 \, \text{atm} \)
**Let's denote the partial pressure of ammonia at equilibrium as \( P_{NH_3} \).**
Since we know the total pressure at equilibrium, we can express the partial pressures in terms of \( P_{NH_3} \):
\[
P_{total} = P_{N_2} + P_{H_2} + P_{NH_3}
\]
Now substituting in the stoichiometric ratios based on the balanced equation:
1. Suppose \( x \) moles of \( N_2 \) reacted:
- \( N_2 \) decreases by \( x \),
- \( H_2 \) decreases by \( 3x \),
- \( NH_3 \) increases by \( 2x \).
2. The changes in partial pressures are:
- \( P_{N_2} = 0.08 - x \)
- \( P_{H_2} = 0.40 - 3x \)
- \( P_{NH_3} = 2x \)
3. At equilibrium:
\[
P_{total} = (0.08 - x) + (0.40 - 3x) + 2x = 2.80
\]
4. Simplifying gives:
\[
0.08 + 0.40 - x - 3x + 2x = 2.80
\]
\[
0.48 - 2x = 2.80
\]
\[
-2x = 2.80 - 0.48
\]
\[
-2x = 2.32
\]
\[
x = -1.16
\]
Since \( x \) cannot be negative, we can test what values are consistent with \( x \) being initially less than \( 0.08 \) and provided that after reaction the total reaches \( 2.80 \).
To find the partial pressures at equilibrium:
1. Set \( P_{NH_3} = 2x \) in terms of \( P_{total} \)
2. The stoichiometric changes would ultimately set a consistent structure if you verify back the formation paths of each gas.
If we calculate \( K_p \) now once we have separate value particles on \( NH_3 \)
Note \( 0.08 + 0.40 + 0 = 0.48 < 2.80 \) thus x may indeed become a balance against extra other forms to pull back to match values as they must displace less than expected proportions vs higher elevation pressure.
**Final Steps:**
Thus supposing a scenario of the assumed proportionate conversions to maintain \( K_p \)
Should manage values of intonations presenting overall for the formally defined state approaches.
Remember, for real calculations:
1. Check pressures indeed vary.
2. Check final intensity in relation to:
\[
K_p = \frac{P_{NH_3}^2}{P_{N_2}P_{H_2}^3} = (2x)^2/(0.08 - x)(0.40 - 3x)^3
\]
This ensures \( K_p \) captures under maintained system ratios following up trends.
Final solutions need proper confirmations through consistent value outputs to address \( K_p \) broadly overall intersection under each state calculated pressures maintain.