Chemistry, 06.12.2019 06:31 btcastongia

Part a what is the numerical value of kc for the following reaction if the equilibrium mixture contains 0.51 m c3h6o, 0.30 m o2, 1.8 m co2, and 2.0 m h2o? c3h6o(g)+4o2(g)⇌3co2(g)+3h2o(g)
a) 2.4 × 101b) 1.1 × 104c) 8.9 × 10-5d) 4.3 × 10-2

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Answer from: princesstn28oqlfir

$1.1\times 10^4$

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_c .

The given balanced equilibrium reaction is,

$C_3H_6O(g)+4O_2(g)\rightleftharpoons 3CO_2(g)+3H_2O(g)$

At eqm. conc. (0.51) M (0.30) M (1.8) M (2.0)M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CO_2]^3\times [H_2O]^3}{[O_2]^4\times [C_3H_6O]^1}$

Now put all the given values in this expression, we get :

$K_c=\frac{(1.8)^3\times (2.0)^3}{(0.30)^4\times (0.51)^1}$

$K_c=1.1\times 10^4$

Thus the value of the equilibrium constant is $1.1\times 10^4$

Answer from: Quest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

Answer from: Quest

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

= 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

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