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3 years ago

For the reaction

1 answer:

8 0

We are asked to find the value of ΔG°rxn from the equilibrium concentrations of the reactants and products. We can use the following formula:

ΔG°rxn = -RTlnK

The value of R = 8.314 J/Kmol, T = 298.15 K and we are given the equilibrium constant Keq = 2.82.

The question provides equilibrium concentrations and then asks to find ΔG°rxn when more of a product is added to the reaction mixture. However, you are asked to find ΔG after the reaction has settled down and reached equilibrium once more. Therefore, we can simply use Keq = 2.82 still and solve for ΔG.

ΔG°rxn = -(8.314 J/Kmol)(298.15 K)(ln(2.82))
ΔG°rxn = -2570 J/mol
ΔG°rxn = -2.57 kJ/mol

Under equilibrium conditions at standard temperature and pressures, the value of ΔG°rxn = -2.57 kJ/mol.

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Reaction Rates

inn [45]

Answer:

PART A 1st order in A and 0th order in B

Part B The reaction rate increases

Explanation:

The rate law of the arbitrary chemical reaction is given by

$-r_A=k\times\left[A\right]^\alpha\times\left[B\right]^\beta\bigm$

Replacing for the data

Expression 1 $0.00639=k\times{0.12}^\alpha\times{0.22}^\beta$

Expression 2 $0.01280=k\times{0.24}^\alpha\times{0.22}^\beta$

Expression 3 $0.00639=k\times{0.12}^\alpha\times{0.11}^\beta$

Making the quotient between the fist two expressions

$\frac{0.00639}{0.01280}=\left(\frac{0.12}{0.24}\right)^\alpha$

Then the expression for $\alpha$

$\alpha=\frac{ln\frac{0.00639}{0.01280}}{ln\frac{0.12}{0.24}}=1\bigm$

Doing the same between the expressions 1 and 3

$\frac{0.00639}{0.00639}=\left(\frac{0.22}{0.11}\right)^\beta$

Then

$\beta=\frac{ln\frac{0.00639}{0.00639}}{ln\frac{0.22}{0.11}}=0\bigm$

This means that the reaction is 1st order respect to A and 0th order respect to B .

By the molecular kinetics theory, if an increment in the temperature occurs, the molecules will have greater kinetic energy and, consequently, will move faster. Thus, the possibility of colliding with another molecule increases. These collisions are necessary for the reaction. Therefore, an increase in temperature necessarily produces an increase in the reaction rate.

4 0

3 years ago

What is true about energy in an ordinary chemical reaction? Energy is... a- created b- destroyed c-not a or b d-used up by react

kap26 [50]

Energy cannot be created or destroyed, so the answer would be c.) not a or b.

Hope this helps!

6 0

3 years ago

A spirit burner used 1.00 g methanol to raise the temperature of 100.0 g water in a metal can from 28.00C to 58.0C. Calculate th

Andreas93 [3]

Complete question:

A spirit burner used 1.00 g methanol to raise the temperature of 100.0 g water in a metal can from 28.00C to 58.0C. Calculate the heat of combustion of methanol in kJ/mol.

Answer:

the heat of combustion of the methanol is 402.31 kJ/mol

Explanation:

Given;

mass of water, $m_w$ = 100 g

initial temperature of water, t₁ = 28 ⁰C

final temperature of water, t₂ = 58 ⁰C

specific heat capacity of water = 4.184 J/g⁰C

reacting mass of the methanol, m = 1.00 g

molecular mass of methanol = 32.04 g/mol

number of moles = 1 / 32.04

= 0.0312 mol

Apply the principle of conservation of energy;

$n\Delta H_{methanol} = Q_{water}\\\\n\Delta H_{methanol} = mc\Delta t\\\\n\Delta H_{methanol} = 100 \times 4.184\times (58-28)\\\\n\Delta H_{methanol} = 12,552 \ J\\\\n\Delta H_{methanol} = 12.552 \ kJ\\\\\Delta H_{methanol} = \frac{12.552}{n} \\\\H_{methanol} = \frac{12.552 \ kJ}{0.0312 \ mol} \\\\\Delta H_{methanol} = 402.31 \ kJ/mol$