What is the total energy change for the following reaction: CO + H2O -> CO2 + H2?

If two atoms are bonded with a bond energy of 87 kJ, what must be done to break the bond?

SSSSS [86.1K]

I would say the energy has to be decreased by 87 kj because the bonding is held together by 87 kj so removing that should prevent the bonding from taking place or reverse it I believe. In other words, a certain amount of energy is required to hold the bond together and in the absence of that energy, the bonding will not take place.

7 0

4 years ago

Please answer ASAP!!!

Karolina [17]

A. Conducting a drug experiment which will harm lab rats

4 0

3 years ago

Read 2 more answers

5.11 g of MgSO₄ is placed into 100.0 mL of water. The water's temperature increases by 6.70°C. Calculate ∆H, in kJ/mol, for the

cupoosta [38]

Answer: Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ

Explanation:

To calculate the entalpy, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed by water = ?

m = mass of water = ${\text {volume of water}}\times {\text {density of water}}=100.0ml\times 1.00g/ml=100.0g$

c = heat capacity of water = 4.186 J/g°C

$\Delta T$ = change in temperature = $6.70^0C$

$q=100.0g\times 4.184J/g^0C\times 6.70^0C=2803.3J=2.8033kJ$

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

The heat absorbed by water will be equal to heat released by $MgSO_4$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 5.11 g

Molar mass = 120 g/mol

Putting values in above equation, we get:

$\text{Moles of }MgSO_4=\frac{5.11g}{120g/mol}=0.042mol$

0.042 moles of $MgSO_4$ releases = 2.8033 kJ

1 mole of $MgSO_4$ releases = $\frac{2.8033 kJ}{0.042}\times 1=66.7kJ$

Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ

3 0

4 years ago

If a solution containing 45.101 g of mercury(II) acetate is allowed to react completely with a solution containing 12.026 g of s

AnnyKZ [126]

Answer:

14.533 grams of solid precipitate of mercury(II) dichromate will form.

Explanation:

$Hg(CH_3COO)_2(aq)+Na_2Cr_2O_7(aq)\rightarrow HgCr_2O_7(s)+2CH_3COONa(aq)$

Moles of mercury(II) acetate = $\frac{45.101 g}{318.70 g/mol}=0.14152 mol$

Moles of sodium dichromate = $\frac{12.026 g}{261.97 g/mol}=0.045906 mol$

According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ of mercury(II) acetate

This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.

According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ of mercury(II) dichromate

Mass of 0.045906 moles of mercury(II) dichromate:

0.045906 mol × 316.59 g/mol = 14.533 g

14.533 grams of solid precipitate of mercury(II) dichromate will form.

3 0

4 years ago

1. According to Le Châtelier's principle, an increase in temperature will shift the equilibrium position toward the products in

sdas [7]

Answer:

1. True

2. Shift it toward the side with lower total mole concentration

3. It will shift toward the product side as there are a fewer number of moles of gas on the product side.

Explanation:

1.

An endothermic reaction is reaction that will absorb energy from the surrounding arena. Increasing temperature will increase the heat of the system. Since the average heat of the surrounding is higher, it will be easier to do an endothermic reaction than exothermic, so this will shift the equilibrium position toward endothermic reaction.

2.

When pressure increase, the molecule will harder to expand. This mean reaction that produces more molecules will be harder to happen since it will take more room and increase the pressure further. This will make the equilibrium shift toward the side with lower total mole concentration since it will help to make more room, thus making the pressure lower.

3.

Remember that only the gas form will contribute to the pressure of the system. In this reaction, there are 3 kinds of gas: nitrogen, hydrogen, and ammonia. Since all form in this reaction gas, you can calculate them all.

On the left side, we have 1 nitrogen and 3 hydrogen so the total is 4.

On the right side, we have 2 ammonia so the total is 2.

When pressure decrease, the equilibrium will shift toward the side with lower total mole concentration, which is the ammonia side. So, the reaction will shift to the product side.

3 0

3 years ago