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3 years ago

I WILL GIVE BRAINLIEST!!!!!

1 answer:

4 0

HNO₃ + H₂S → S + NO + H₂<span>O

Assign Oxidation Number:

L.H.S R.H.S
N in HNO</span>₃ = +5 +2 = N in NO
S in H₂S = -2 0 = S in S

Write Half cell Reactions:
Reduction Reaction:
3e⁻ + HNO₃ → NO -------(1)
Oxidation Reaction:
H₂S → S + 2e⁻ -------(2)

Multiply eq. 1 with 2 and eq. 2 with 3 to balance electrons.

6e⁻ + 2 HNO₃ → 2 NO

3 H₂S → 3 S +  6e⁻
Cancel e⁻s,
______________________________

2 HNO₃ + 3 H₂S → 2 NO + 3 S + H₂O

Balance Oxygen Atoms by multiplying H₂O with 4, Hydrogen atoms will automatically get balance.

2 HNO₃ + 3 H₂S → 2 NO + 3 S + 4H₂O

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2/3

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Equilibrium concentrations of the gases are

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Explanation:

We are given that for the equilibrium

$2H_2S\rightleftharpoons 2H_2(g)+S_2(g)$

$k_c=9.0\times 10^{-8}$

Temperature, $T=700^{\circ}C$

Initial concentration of

$H_2S=0.30M$

$H_2=0.30 M$

$S_2=0.150 M$

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles of reactant reduced and form product

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Equilibrium constant

$K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}$

Substitute the values

$9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}$

By solving we get

$x\approx 0.148$

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