312213
Apr16-09, 01:09 PM
1. The problem statement, all variables and given/known data
H2 (g) + Cl2(g) \Updownarrow 2HCl (g) Kp = 2.5 × 1033
NH3(g) + HCl(g) \Updownarrow NH4Cl(s) Kp = 2.1 × 1015
N2(g) + 4H2(g) +Cl2 \Updownarrow 2NH4Cl(s) Kp = 3.9 × 1070
Determine the Kp for N2(g) + 3H2 (g) \Updownarrow 2NH3(g).
2. Relevant equations
Don't know/none
3. The attempt at a solution
In order to get to N2(g) + 3H2 (g) \Updownarrow 2NH3(g), I would have to multiple/flip equations so that they result in the desired reaction set.
-(H2 (g) + Cl2(g) \Updownarrow 2HCl (g) Kp = 2.5 × 1033)
-2(NH3(g) + HCl(g) \Updownarrow NH4Cl(s) Kp = 2.1 × 1015)
N2(g) + 4H2(g) +Cl2 \Updownarrow 2NH4Cl(s) Kp = 3.9 × 1070
These would cancel out to the desired reaction.
In Hess's Law, I understand that multiplying a step would mean its enthalpy gets multiplied by that number. If I flip a step, its enthalpy would inverse its sign.
In voltage calculation from standard reduction potentials, reversing the sign would inverse the potential for the step but multiplying the step does not affect the potential.
Originally I would just follow Hess's Law to calculate but I never did this for Kp and so I'm not sure the answer would be correct. How do I approach this problem and solving for Kp.
H2 (g) + Cl2(g) \Updownarrow 2HCl (g) Kp = 2.5 × 1033
NH3(g) + HCl(g) \Updownarrow NH4Cl(s) Kp = 2.1 × 1015
N2(g) + 4H2(g) +Cl2 \Updownarrow 2NH4Cl(s) Kp = 3.9 × 1070
Determine the Kp for N2(g) + 3H2 (g) \Updownarrow 2NH3(g).
2. Relevant equations
Don't know/none
3. The attempt at a solution
In order to get to N2(g) + 3H2 (g) \Updownarrow 2NH3(g), I would have to multiple/flip equations so that they result in the desired reaction set.
-(H2 (g) + Cl2(g) \Updownarrow 2HCl (g) Kp = 2.5 × 1033)
-2(NH3(g) + HCl(g) \Updownarrow NH4Cl(s) Kp = 2.1 × 1015)
N2(g) + 4H2(g) +Cl2 \Updownarrow 2NH4Cl(s) Kp = 3.9 × 1070
These would cancel out to the desired reaction.
In Hess's Law, I understand that multiplying a step would mean its enthalpy gets multiplied by that number. If I flip a step, its enthalpy would inverse its sign.
In voltage calculation from standard reduction potentials, reversing the sign would inverse the potential for the step but multiplying the step does not affect the potential.
Originally I would just follow Hess's Law to calculate but I never did this for Kp and so I'm not sure the answer would be correct. How do I approach this problem and solving for Kp.