Decompose to water and oxygen

In summary, "Decomposing to water and oxygen" is a process known as water electrolysis, which involves passing an electric current through water to split it into hydrogen and oxygen molecules. The main products of this process are hydrogen gas and oxygen gas, which can be used as clean and renewable sources of energy. The chemical equation for this process is 2H2O(l) → 2H2(g) + O2(g), and the rate of decomposition can be affected by factors such as electrolyte concentration, voltage and current, and electrode surface area.
  • #1
yuuri14
28
0
hydrogen peroxide can decompose to water and oxygen by following reaction:
2H2 O2 (l) = 2H2 O(l) + O2(g) deltaH= -196kJ

CALCULATE THE VALUE OF q when 5.00g of H2 O2(l) decomposes at constant pressure.

my answer is -14.4J...can some please tell me how to do this again because I forgot what step I took and it have a similar question on another homework
 
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  • #2
delta H is in terms of kJ per mole of equation; how many moles of H2O2 are there in 5.00g and how many 'moles of equation' are there?
 
  • #3


To calculate the value of q, we can use the equation q = mCΔT, where q is the heat energy, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

First, we need to determine the molar mass of H2O2 by adding the atomic masses of each element (2 hydrogen atoms + 2 oxygen atoms) = 2(1.008 g/mol) + 2(15.999 g/mol) = 34.014 g/mol.

Next, we need to convert the given mass of H2O2 (5.00g) to moles by dividing it by the molar mass: 5.00g / 34.014 g/mol = 0.147 mol.

Since the reaction produces 2 moles of H2O for every 1 mole of H2O2, we know that 0.147 mol of H2O2 will produce 2(0.147) = 0.294 mol of H2O.

Now, we can use the given information about the enthalpy change (deltaH = -196 kJ) to calculate the heat energy released during the reaction. We need to convert the given enthalpy change to joules by multiplying by 1000 (1 kJ = 1000 J) and then multiplying by the number of moles produced in the reaction: -196 kJ * 1000 J/kJ * 0.294 mol = -57,624 J.

Finally, we can plug in the values we calculated into the equation q = mCΔT and solve for q. We know the mass of H2O produced (0.294 mol * 18.015 g/mol = 5.29 g), the specific heat capacity of water (4.18 J/g°C) and the change in temperature (the reaction is assumed to occur at constant pressure, so ΔT = 0). Therefore, q = 5.29 g * 4.18 J/g°C * 0 = 0 J.

Therefore, the value of q when 5.00 g of H2O2 decomposes at constant pressure is 0 J. Your previous answer of -14.4 J may have been due to rounding errors or a different method of calculation. However, it is important to double check your calculations and units to ensure accuracy.
 

What is the process of "Decomposing to water and oxygen"?

The process of "Decomposing to water and oxygen" is known as water electrolysis. It involves passing an electric current through water, causing it to split into hydrogen and oxygen molecules.

What are the main products of "Decomposing to water and oxygen"?

The main products of "Decomposing to water and oxygen" are hydrogen gas and oxygen gas. The hydrogen gas is collected at the negative electrode (cathode) and the oxygen gas is collected at the positive electrode (anode).

What is the purpose of "Decomposing to water and oxygen"?

The purpose of "Decomposing to water and oxygen" is to produce hydrogen and oxygen, which can be used as clean and renewable sources of energy. Hydrogen can be used in fuel cells to power vehicles and generate electricity, while oxygen can be used in medical and industrial applications.

What is the chemical equation for "Decomposing to water and oxygen"?

The chemical equation for "Decomposing to water and oxygen" is 2H2O(l) → 2H2(g) + O2(g). This equation shows that two molecules of water (H2O) are broken down into two molecules of hydrogen gas (H2) and one molecule of oxygen gas (O2).

What are the factors that affect the rate of "Decomposing to water and oxygen"?

The rate of "Decomposing to water and oxygen" can be affected by several factors, including the concentration of the electrolyte, the voltage and current applied, and the surface area of the electrodes. Higher concentrations of electrolyte and higher voltages and currents will result in a faster rate of decomposition, while a larger surface area of the electrodes will increase the rate of hydrogen and oxygen production.

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