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[ 2 mathrm(H)_(2) mathrm(~S)(g) leftharpoons 2 mathrm(H)_(2)(g)+mathrm(S)_(2)(g) ] A rigid reaction vessel contains mathrm(H)_(2) mathrm(~S)(g) in equilibrium with mathrm(H)_(2)(g) and mathrm(S)_(2)(g) at 1200 mathrm(~K) , as represented by the equation above. The partial pressures of the gases at equilibrium arif shown in the table below. Gas & }(c) Equilibrium partial pressure (atm) mathrm(H)_(2) mathrm(~S) & 1.2 mathrm(H)_(2) & 0.21 mathrm(~S)_(2) & 0.016 Calculate the value of the equilibrium constant, K_(p) , for the reaction at 1200 mathrm(~K) .

Question

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2 mathrm(H)_(2) mathrm(~S)(g) leftharpoons 2 mathrm(H)_(2)(g)+mathrm(S)_(2)(g)
]
A rigid reaction vessel contains mathrm(H)_(2) mathrm(~S)(g) in equilibrium with mathrm(H)_(2)(g) and mathrm(S)_(2)(g) at 1200 mathrm(~K) , as represented by the equation above. The partial pressures of the gases at equilibrium arif shown in the table below.

Gas & }(c)
Equilibrium partial 
pressure (atm)
 
 mathrm(H)_(2) mathrm(~S) & 1.2 
 mathrm(H)_(2) & 0.21 
 mathrm(~S)_(2) & 0.016

Calculate the value of the equilibrium constant, K_(p) , for the reaction at 1200 mathrm(~K) .

[ 2 mathrm(H)_(2) mathrm(~S)(g) leftharpoons 2 mathrm(H)_(2)(g)+mathrm(S)_(2)(g) ] A rigid reaction vessel contains mathrm(H)_(2) mathrm(~S)(g) in equilibrium with mathrm(H)_(2)(g) and mathrm(S)_(2)(g) at 1200 mathrm(~K) , as represented by the equation above. The partial pressures of the gases at equilibrium arif shown in the table below. Gas & }(c) Equilibrium partial pressure (atm) mathrm(H)_(2) mathrm(~S) & 1.2 mathrm(H)_(2) & 0.21 mathrm(~S)_(2) & 0.016 Calculate the value of the equilibrium constant, K_(p) , for the reaction at 1200 mathrm(~K) .

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EthanElite · Tutor for 8 years

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#Explanation<br />The equilibrium constant, \( K_p \), is a measure of the ratio of the concentrations of the products to the reactants at equilibrium, each raised to the power of their stoichiometric coefficients in the balanced chemical equation. For the given reaction, the equilibrium constant \( K_p \) is given by:<br /><br />\[ K_p = \frac{[H_2]^2[S_2]}{[H_2S]^2} \]<br /><br />where [H_2S], [H_2], and [S_2] represent the equilibrium partial pressures of \( H_2S \), \( H_2 \), and \( S_2 \) respectively.<br /><br />Given the equilibrium partial pressures of the gases:<br /><br />\[ [H_2S] = 1.2 \, atm \]<br />\[ [H_2] = 0.21 \, atm \]<br />\[ [S_2] = 0.016 \, atm \]<br /><br />we can substitute these values into the equation for \( K_p \) to calculate its value.<br /><br />#Answer<br />Substituting the given values into the equation for \( K_p \), we get:<br /><br />\[ K_p = \frac{(0.21)^2 \times 0.016}{(1.2)^2} \]<br /><br />Calculating the above expression, we find that \( K_p \) is approximately 0.000588.
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