A 1.50 L sample of gaseous HI having a density of
0.0101 gcm3 is heated at 410 °C. As time passes, the
HI decomposes to gaseous H2 and I2. The rate law is
-Δ3HI4Δt = k3HI42, where k = 0.0311M ~ min2 at
410 °C.
(b) What is the partial pressure of H2 after a reaction time
of 8.00 h?

Question

A 1.50 L sample of gaseous HI having a density of 
0.0101 g>cm3 is heated at 410 °C. As time passes, the 
HI decomposes to gaseous H2 and I2. The rate law is 
-Δ3HI4>Δt = k3HI42, where k = 0.031>1M ~ min2 at 
410 °C. 
(b) What is the partial pressure of H2 after a reaction time 
of 8.00 h?

Accepted Answer

Hello everyone today. We have the following question when heated at a 520 degrees Celsius. A sample of 2.75 liters of nitric oxide gas, having a density of 0.134 g from middle leader gradually breaks down into nitrogen gas and oxygen gas. If it's rate law is equal to the following Where we have the K value equal to the following at 520°C determined the partial pressure of nitrogen gas after four hours. So before we do that, we have to find a couple of things. The first thing you must find is the moles of our nitric oxide And we can do that by starting with our volume. We started with 2.75 l and we're going to convert that into milliliters by using the conversion factor that there are 1000 mL in one liter. And then we're going to multiply by the density, which was 0.00134 g per mila leader. And then lastly, we're gonna multiply by the molar mass, which The molar mass states that for one mole of our nitric oxide, we're going to get this from the periodic table. We have 30.01g of nitric oxide. And so our units will all cancel out Except for our moles. We're gonna be left with 0.1228 moles of our nitric oxide from that we can find the initial concentration Of our nitric oxide. And that's going to be in terms of moles per liter. So we're gonna take our 0.1, moles And divide that by our leaders or volume which was 2.75 L. And this is going to give us 0. moller. And that's our initial concentration. Next we need to find the rate. So our rate will be equal to some constant K. Times our concentration of nitric oxide raised to the power of two. So we're gonna use the rate law that was given to us in the question. In that rate law, one over our nitric oxide at a given time is equal to our constant times our time plus one over The nitric oxide at T equals zero. So this is gonna look like we're gonna have our 0.021 over Moeller per minute. And this is given to us in the question stem. What we're gonna do is multiply this by our four hours And then convert those hours into minutes and say that we have 60 minutes per one hour. And then at that two hour one over the concentration of nitric oxide. Our initial concentration which is 0.04465 molar. And when we do that we're going to get our concentration equal to zero Or we're gonna get our concentration equal to .4355 over Moller. We're gonna need to do, We're gonna need to divide this or we're going to need to do one divided by that number. To give us 0.03645 molar. So next we can actually write this equation out or this reaction out, we're gonna have our nitric oxide breaking down into nitrogen gas and oxygen gas. Now we only have one nitrogen, one oxygen on the left, however, we have two of each on the right, so we simply use it added to coefficient to the nitric oxide on the left there. So what we can do from this is we can find our concentration of our nitrogen gas That's going to be 1/ of our concentration of our nitric oxide at a given time or initial minus the concentration of nitric oxide at a given time. T And so when we do this, we get one half is equal to 0.4465 molar minus 0.3645 molar. And this gives us 0.0041 molar. For a concentration of extreme gas. Of course our temperature cannot remain in C. It has to be in Kelvin. So we're going to take that Degrees C and add to 73.15 to it to get 793.15 Kelvin. Now that we have that we can plug this into our ideal gas law. And so we're gonna have to rearrange this actually. And so to do that, we're gonna have our pressure is equal to the moles, terms are gas constant, times the temperature over our volume. And this is for the partial pressure. We plug in our values, we're going to get 0.0041 moles per liter. That was the polarity that we have. What we're gonna do is we're gonna multiply by the gas constant 0.8 to Leaders Times atmospheres, moles, times kelvin And then by the temperature, which is 793 Kelvin. And when we do that, we get 0.2668 atmospheres as our partial pressure and our final answer. And with that we have answered the question overall. I hope this helped. And until next time.

A 1.50 L sample of gaseous HI having a density of 0.0101 g>cm3 is heated at 410 °C. As time passes, the HI decomposes to gaseous H2 and I2. The rate law is -Δ3HI4>Δt = k3HI42, where k = 0.031>1M ~ min2 at 410 °C. (b) What is the partial pressure of H2 after a reaction time of 8.00 h?

Verified Solution

Key Concepts

Ideal Gas Law

Rate Law and Reaction Kinetics

Partial Pressure