<u>Answer:</u> The equilibrium concentration of
is 1.285 M.
<u>Explanation:</u>
The chemical equation for the decomposition of phosphorus pentachloride follows:

The expression for equilibrium constant is given as:
![K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
We are given:

![[PCl_3]=0.18M](https://tex.z-dn.net/?f=%5BPCl_3%5D%3D0.18M)
![[Cl_2]=0.30M](https://tex.z-dn.net/?f=%5BCl_2%5D%3D0.30M)
The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.
Putting values in above equation, we get:
![0.042=\frac{0.18\times 0.30}{[PCl_5]}](https://tex.z-dn.net/?f=0.042%3D%5Cfrac%7B0.18%5Ctimes%200.30%7D%7B%5BPCl_5%5D%7D)
![[PCl_5]=1.285](https://tex.z-dn.net/?f=%5BPCl_5%5D%3D1.285)
Hence, the equilibrium concentration of
is 1.285 M.
We know that acids have a pH of under 7.
We also need to:
Set up an ICE table for the chemical reaction. Solve for the concentration of H3O+ using the equation for pH Use the concentration of H3O+ to solve for the concentrations of the other products and reactants.
Answer:
Hi.. U should do like this
Cu + S - - - >Cu2S
Answer
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Answer:
Explanation:
Ba(s) + Mn²⁺ (aq,1M) → Ba²⁺ (aq,1M) + Mn(s)
Ba⁺²(aq) +2e → Ba(s) , E° = −2.90 V
Mn⁺²(aq) +2e → Mn(s), E⁰ =0.80 V
Anode reaction :
Ba(s) → Ba⁺²(aq) +2e E° = −2.90 V
Cathode reaction :
Mn⁺²(aq) +2e → Mn(s) E⁰ =0.80 V
Cell potential = Ecathode - Eanode
Ecell = .80 - ( - 2.90 )
Ecell = 3.7 V .
equilibrium constant ( K ) :
Ecell = .059 log K / n
n = 2
3.7 = .059 log K / 2
log K = 125.42
K = 2.63 x 10¹²⁵ .
Free energy change :
ΔG = - n F Ecell
= - 2 x 96500 x 3.7
= 714100 J
= 7.141 x 10⁵ J .