Please help!

N2H4 is decomposed in a closed container according to this reaction: N2H4 → N2 + 2H2 . If the temperature is constant and the initial pressure of N2H4 is 800mm, what will the pressure in millimeters be after the decomposition to N2 + 2H2?

P V = n R T

You have three times as many molecules or mols
new n = 3 * original n
V, R and T the same
so new P = 3*old P

To determine the pressure after the decomposition of N2H4, we need to use the ideal gas law. The ideal gas law equation is:

PV = nRT

Where:
P = pressure
V = volume
n = moles of gas
R = ideal gas constant
T = temperature

In this case, the volume and temperature remain constant. So, we can rewrite the ideal gas law equation as:

P1/n1 = P2/n2

Where:
P1 = initial pressure of N2H4
n1 = initial moles of N2H4
P2 = final pressure after decomposition
n2 = final moles of gas (N2 + 2H2)

The balanced chemical equation gives us the stoichiometric ratio between N2H4 and the products N2 and H2. From the equation, we can see that 1 mole of N2H4 decomposes to form 1 mole of N2 and 2 moles of H2.

So, n2 = nN2 + nH2 = n1 (moles of N2H4)

Given that the initial pressure of N2H4 is 800 mmHg, we can use these values to calculate the final pressure:

P1/n1 = P2/n2

800 mmHg / n1 = P2 / (n1 + n1*2)

Simplifying the equation:

800 mmHg = P2 / 3

P2 = 800 mmHg * 3

P2 = 2400 mmHg

Therefore, the pressure after the decomposition of N2H4 to N2 + 2H2 will be 2400 mmHg.