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3 years ago

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For an exothermic reaction at equilibrium, how will increasing the temperature affect Keq?

1 answer:

Sonja [21]3 years ago

8 0

Question:

<em>For an exothermic reaction at equilibrium, how will increasing the temperature affect Keq?</em>

Answer:

<em>The reaction will proceed towards the liquid phase. Heat is on the reactant side of the equation. Lowering temperature will shift equilibrium left, creating more liquid water. A reaction that is exothermic releases heat, while an endothermic reaction absorbs heat.</em>

<em>If you increase the temperature, the position of equilibrium will move in such a way as to reduce the temperature again. It will do that by favouring the reaction which absorbs heat. In the equilibrium, that will be the back reaction because the forward reaction is exothermic.</em>

Hope this helps, have a good day. c;

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Assume the hydrolysis of ATP proceeds with ΔG′° = –30 kJ/mol. ATP + H2O → ADP + Pi Which expression gives the ratio of ADP to AT

Andru [333]

Answer:

$6.14\cdot 10^{-6}$

Explanation:

Firstly, write the expression for the equilibrium constant of this reaction:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

$\Delta G^o = -RT ln K_{eq}$

From here, rearrange the equation to solve for K:

$K_{eq} = e^{-\frac{\Delta G^o}{RT}}$

Now we know from the initial equation that:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Let's express the ratio of ADP to ATP:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}$

Substitute the expression for K:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}$

Now we may use the values given to solve:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}$

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2 years ago

Calculate ΔG o for the following reaction at 25°C: 3Mg(s) + 2Al3+(aq) ⇌ 3Mg2+(aq) + 2Al(s) Enter your answer in scientific notat

larisa [96]

Answer:

-3.7771 × 10² kJ/mol

Explanation:

Let's consider the following equation.

3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)

We can calculate the standard Gibbs free energy (ΔG°) using the following expression.

ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)

where,

n: moles

ΔG°f(): standard Gibbs free energy of formation

p: products

r: reactants

ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))

ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)

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In one of the routine analyses, Dr Entropy found that the concentration of phosphate had increased from 0.0090 mg/L to 0.220 mg/

wariber [46]

Answer:

The final volume should be 22 mL

Explanation:

For this problem, we will use the dilution equation:

C1*V1 = C2*V2

<u>Step 1</u>: Data given

with C1 = the initial concentration C1 = 0.220 mg/L

with V1 = the initial volume = 10 mL = 10 * 10^-3 L

with C2 = the final concentration = 0.100 mg/L

with V2 = the final volume = TO BE DETERMINED

<u>Step 2</u>: Calculating the final volume

C1*V1 = C2*V2

0.220 mg/L * 10*10^-3 L = 0.100 mg/L * V2

V2 = (0.220 mg/L * 10*10^-3 L) / 0.100 mg/L

V2 =0.022 L = 22 mL

The final volume should be 22 mL

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