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SOVA2 [1]
3 years ago
9

The reaction 2PH3(g)+As2(g)⇌2AsH3(g)+P2(g) has Kp=2.9×10−5 at 873 K. At the same temperature, what is Kp for each of the followi

ng reactions?Parts A, B, and CPart A:2AsH3(g)+P2(g)⇌2PH3(g)+As2(g)Part B:6PH3(g)+3As2(g)⇌3P2(g)+6AsH3(g)Part C:2P2(g)+4AsH3(g)⇌2As2(g)+4PH3(g)
Chemistry
1 answer:
slavikrds [6]3 years ago
8 0

Answer:

Part A

Kp = 3.4 x 10⁴

Part B

Kp = 2.4 x 10⁻¹⁴

Part C

Kp = 1.2 x 10⁹

Explanation:

2PH₃(g) + As₂(g)  ⇌ 2 AsH₃(g) + P₂(g)  Kp = 2.9 x 10⁻⁵

Kp = [AsH₃]²[P₂]/[PH₃]²[As]  = 2.9 x 10⁻⁵

Part A

it is the inverse of the equilibrium given

Kp(A) = 1/ Kp = 1 / 2.9 x 10⁻⁵ = 3.4 x 10⁴

Part B

Is the equilibrium where the coefficients have been multiplied by 3,

Kp(B) = ( Kp )³ = ( 2.9 x 10⁻⁵ )³ = 2.4 x 10⁻¹⁴

Part C

This is the  reverse equilibrium multipled by 2.

Kp(C) = ( 1/Kp)² = ( 1/ 2.9 x 10⁻⁵ )² = 1.2 x 10⁹

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Answer:

Both Option (I) and Option (III)

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4 years ago
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4 0
3 years ago
Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a
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Answer : The correct option is, (b) +0.799 V

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The values of standard reduction electrode potential of the cell are:

E^0_{[H^{+}/H_2]}=+0.00V

E^0_{[Ag^{+}/Ag]}=+0.799V

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

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Reaction at cathode (reduction) : Ag^{+}+1e^-\rightarrow Ag    

The balanced cell reaction will be,  

H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag

Now we have to calculate the standard electrode potential of the cell.

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}

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Therefore, the standard cell potential will be +0.799 V

4 0
3 years ago
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