4 years ago

(?)Na2HPO4 → (?)Na4P2O7 + (?)H2O

Chemistry

1 answer:

Akimi4 [234]4 years ago

7 0

Answer: 2,1,1

Explanation: (2)Na2HPO4=(1)Na4P2O7+(1)H20

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Given that Kp = 3.5 x 10-4 for the reaction 2 CO(g) <=> C(graphite) + CO2(g), what is the partial pressure of CO2(g) at eq

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The given chemical equation follows:

$2CO(g)\rightleftharpoons C\text{ (graphite)}+CO_2(g)$

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At eqllm: 4.00-2x x x

The expression of $K_p$ for above reaction follows:

$K_p=\frac{p_{CO_2}}{(p_{CO})^2}$

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We are given:

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Putting values in above expression, we get:

$3.5\times 10^{-4}=\frac{x}{(4-2x)^2}\\\\x=0.0056,718.28$

Neglecting the value of x = 718.28 because equilibrium pressure cannot be greater than initial pressure

Partial pressure of $CO_2$ = 0.0056 atm

Hence, the partial pressure of carbon dioxide at equilibrium is 0.0056 atm

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