Of the choices below, which is true for the relationship shown? it is ka for the acid h3p2o72â. it is kb for the acid h3p2o72â.
2 answers:
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The question is incomplete, the complete question is:
At a certain temperature, the equilibrium constant for the following reaction is 0.74:
Suppose a 49.0 L reaction vessel is filled with 1.5 mol of and 1.5 mol of NO. What can you say about the composition of the mixture in the vessel at equilibrium?
A. There will be very little and NO.
B. There will be very little
C. Neither of the above is true.
<u>Answer:</u> The correct option is B. there will be very little
<u>Explanation:</u>
We are given:
Initial moles of = 1.5 moles
Initial moles of NO = 1.5 moles
Volume of vessel = 49 L
As, moles of reactants and moles of products are equal, the volume term will not appear in the equilibrium constant expression.
Equilibrium constant for the reaction = 0.74
For the given chemical equation:
Initial: 1.5 1.5 -
At eqllm: 1.5-x 1.5-x 2x
The expression of equilibrium constant for the above reaction:
Putting values in above equation, we get:
Neglecting the negative value of equilibrium constant because concentration cannot be negative
Equilibrium concentration of = (1.5 - x) = (1.5 - 0.451) = 1.049 moles
Equilibrium concentration of NO = (1.5 - x) = (1.5 - 0.451) = 1.049 moles
Equilibrium concentration of = 2x =
There are 3 possibilities:
- If
, the reaction is reactant favored
- If
, the reaction is product favored.
- If
, the reaction is in equilibrium.
Here, the value of , which is less than 1, therefore the reaction is reactant favored and we can say that there will be very little
.
Hence, the correct option is B. there will be very little
Answer:
−2399.33 kJ
Explanation:
If NH₄NO₃ reacts with fuel oil to give a ΔH of -7198 for every 3 moles of NH₄NO₃
What is the enthalpy change for 1.0 mole of NH₄NO₃ in this reaction
∴ For every 1 mole, we will have of the total enthaply of the 3 moles
so, to determine the 1 mole; we have:
= −2399.33 kJ
∴ the enthalpy change for 1.0 mole of NH₄NO₃ in this reaction = −2399.33 kJ