All categories

3 years ago

For an endothermic reaction, how will the value for Keq change when the temperature is increased?

Chemistry

2 answers:

Nataly [62]3 years ago

4 0

I put the answer C: Keq will increase, on PLATO. Hope this works for you!

Vlada [557]3 years ago

3 0

Answer:

C. Keq will increase

Explanation:

Endothermic means the reaction will absorb heat to occur, hence an increase in temperature will favor the reaction, increasing the formation of products.

Keq is the equilibrium constant of the reaction, and it can be defined as:

$Keq = \frac{concentration of products}{concentration of reactants}$

In an endothermic reaction the equilibrium will be shift towards the formation of products when the temperature is increased, the concentration of products will increase (in relation to the concentration of reactants) ; so as a result the equilibrium constant Keq will also increase.

You might be interested in

The walls in your home have 25mm of expanded polystyrene (R-value 3.96) and 25mm of rigid urethane foam (R-value 7.50). What is

Finger [1]

Answer:

3.6 per inch

EPS insulation features an average r-value of 3.6 per inch.

In addition to providing energy efficiency, EPS without film facers has a perm rating of up to 5.0.

Explanation:

hope this helps

7 0

3 years ago

The use of evaporation and condensation to separate the partsof a mixture is called _____

Keith_Richards [23]

Answer: Evaporation Condensation is used to seperate the parts of a mixture

Explanation: This is because it is the first order of the distilation process

5 0

2 years ago

After fertilization, female cones become very: sticky light O hard

S_A_V [24]

Answer:

light

Explanation:

sorry don't now this one

8 0

3 years ago

N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol

docker41 [41]

Answer:

Explanation:

Recall that ΔH is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:

$\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)$

Therefore, from the chemical equation, we have that:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} + \Delta H^\circ_f \text{ H$_2$O} \right] -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}$

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}$

In conclusion, our answer is A.

5 0

2 years ago

A positive test with 2,4 dinitrophenylhydrazine could indicate the presence of which of the following functional groups ?

IrinaK [193]

6 &&;$;$//&283$-$-&//

8 0

3 years ago