Calculate 6890/1230000 and conclude using the correct scientific figure.

For the simple decomposition reactionAB(g)→ A(g) + B(g)Rate =k[AB]2 and k=0.2 L/mol*s . How long will it takefor [AB] to reach 1

mixer [17]

Answer:

6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.

Explanation:

$Rate = k[AB]^2$

The order of the reaction is 2.

Integrated rate law for second order kinetic is:

$\frac{1}{[A_t]} = \frac{1}{[A]_0}+kt$

Where, $[A_0]$ is the initial concentration = 1.50 mol/L

$[A_t]$ is the final concentration = 1/3 of initial concentration = $\frac{1}{3}\times 1.50\ mol/L$ = 0.5 mol/L

Rate constant, k = 0.2 L/mol*s

Applying in the above equation as:-

$\frac{1}{0.5} = \frac{1}{1.50}+0.2t$

$\frac{1}{1.5}+0.2x=\frac{1}{0.5}$

$t = 6.66\ s$

6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.

5 0

3 years ago

Please help! it’s due date is in a few minutes

Licemer1 [7]

Answer:

40'0'

Explanation:

X is smaller than |-40|

7 0

3 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

Answer: The $\Delta G$ for the reaction is 54.425 kJ/mol

Explanation:

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago

A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions

Mars2501 [29]

Answer:

Qm = -55.8Kj/mole

Explanation:

NaOH(aq) + HNO₃(aq) => NaNO₃(aq) + H₂O(l)

Qm = (mc∆T)water /moles acid

Given => 100ml(0.300M) NaOH(aq) + 100ml(0.300M)HNO₃(aq)

=> 0.03mole NaOH(aq) + 0.03mole HNO₃(aq)

=> 0.03mole NaNO₃(aq) + 0.03mole H₂O(l)

ΔH⁰rxn = [(200ml)(1.00cal/g∙°C)(37 – 35)°C]water / 0.03mole HNO₃

= 13,333 cal/mole x 4.184J/cal = 55,787J/mol = 55.8Kj/mole (exothermic)*

Heat of reactions comes from formation of H-Oxy bonds on formation of water of reaction and heats the 200ml of solvent water from 35⁰C to 37⁰C.

4 0

4 years ago

Read 2 more answers

Answer in detail?? What precautions would you take to protect yourself from an earthquake, when you are at home and when you are

Dmitry [639]

Answer:

If you're indoors, stay inside. If you're outside, stay outside. If you're indoors, stand against a wall near the center of the building, stand in a doorway, or crawl under heavy furniture (a desk or table). Stay away from windows and outside doors.

4 0

3 years ago

Read 2 more answers