If we start the reaction with partial
pressure of A and B as 4atm each what will
be the equibirillium partial pressure & the
total pressure?
Answers
Answer:
dont know the answer
Explanation:
Answer:
The equilibrium constant, K_\text pK
p
K, start subscript, start text, p, end text, end subscript, describes the ratio of product and reactant concentrations at equilibrium in terms of partial pressures.
For a gas-phase reaction, \text{aA}(g)+\text{bB}(g) \leftrightharpoons \text{cC}(g)+\text{dD}(g)aA(g)+bB(g)⇋cC(g)+dD(g)start text, a, A, end text, left parenthesis, g, right parenthesis, plus, start text, b, B, end text, left parenthesis, g, right parenthesis, \leftrightharpoons, start text, c, C, end text, left parenthesis, g, right parenthesis, plus, start text, d, D, end text, left parenthesis, g, right parenthesis, the expression for K_\text pK
p
K, start subscript, start text, p, end text, end subscript is
K_\text p =\dfrac{(\text P_{\text C})^c (\text P_{\text D})^d}{(\text P_{\text A})^a (\text P_{\text B})^b}K
p
=
(P
A
)
a
(P
B
)
b
(P
C
)
c
(P
D
)
d
K, start subscript, start text, p, end text, end subscript, equals, start fraction, left parenthesis, start text, P, end text, start subscript, start text, C, end text, end subscript, right parenthesis, start superscript, c, end superscript, left parenthesis, start text, P, end text, start subscript, start text, D, end text, end subscript, right parenthesis, start superscript, d, end superscript, divided by, left parenthesis, start text, P, end text, start subscript, start text, A, end text, end subscript, right parenthesis, start superscript, a, end superscript, left parenthesis, start text, P, end text, start subscript, start text, B, end text, end subscript, right parenthesis, start superscript, b, end superscript, end fraction
K_\text pK
p
K, start subscript, start text, p, end text, end subscript is related to the equilibrium constant in terms of molar concentration, K_\text cK
c
K, start subscript, start text, c, end text, end subscript, by the equation below:
K_\text p = K_\text c(\text{RT})^{\Delta \text n}K
p
=K
c
(RT)
Δn
K, start subscript, start text, p, end text, end subscript, equals, K, start subscript, start text, c, end text, end subscript, left parenthesis, start text, R, T, end text, right parenthesis, start superscript, delta, start text, n, end text, end superscript
where \Delta \text nΔndelta, start text, n, end text is
\Delta \text n=\text{mol of product gas}-\text{mol of reactant gas}Δn=mol of product gas−mol of reactant gasdelta, start text, n, end text, equals, start text, m, o, l, space, o, f, space, p, r, o, d, u, c, t, space, g, a, s, end text, minus, start text, m, o, l, space, o, f, space, r, e, a, c, t, a, n, t, space, g, a, s, end text
Introduction: a short review of equilibrium and K_\text cK
c
K, start subscript, start text, c, end text, end subscript
When a reaction is at equilibrium, the forward reaction and reverse reaction have the same rate. The concentrations of the reaction components stay constant at equilibrium, even though the forward and backward reactions are still occurring.
Explanation:
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