## Ch 8 #53 Part b

Michelle Dong 1F
Posts: 110
Joined: Fri Sep 29, 2017 7:04 am

### Ch 8 #53 Part b

The reaction of 1.40 g of carbon monoxide with excess water vapor to produce carbon dioxide and hydrogen gases in a bomb calorimeter causes the temperature of the calorimeter assembly to rise from 22.113 C to 22.799 C. The calorimeter assembly is known to have a total heat capacity of 3.00 kJ/C.

(b) Calculate the internal energy change, delta U for the reaction of 1.00 mol CO (g).

404995677
Posts: 82
Joined: Fri Sep 29, 2017 7:07 am

### Re: Ch 8 #53 Part b

I also have a question about why we use q=C(delta)T instead of q=nC(delta)T

RenuChepuru1L
Posts: 58
Joined: Thu Jul 27, 2017 3:00 am

### Re: Ch 8 #53 Part b

why is it c(deltaT) divided by moles and not c(deltaT)(moles)

Rohan Chaudhari- 1K
Posts: 32
Joined: Fri Sep 29, 2017 7:06 am

### Re: Ch 8 #53 Part b

The reason it is negative is because the temperature raises, meaning that the reaction had to release energy. The reason it is just q=C*deltaT is because it is the equation for the calorimeter, which does not include mass.

Diane Bui 2J
Posts: 61
Joined: Sat Jul 22, 2017 3:00 am

### Re: Ch 8 #53 Part b

Hi RenuChepuru1L,
The heat of the reaction is equal to the negative heat of the calorimeter because the calorimeter is gaining heat from the reaction. Hence to calculate heat gained by the calorimeter, we use C(delta T). The question asks for the internal energy change for the reaction of 1.00 mol CO. We are given 1.40 g of CO or 0.04998 mol CO. To find out the heat for 1 mol of CO, we divide the value of C(delta T) by 0.04998 mol CO to get -41.2 kJ/mol CO.

Timothy Kim 1B
Posts: 62
Joined: Fri Sep 29, 2017 7:04 am

### Re: Ch 8 #53 Part b

Diane Bui 2J wrote:Hi RenuChepuru1L,
The heat of the reaction is equal to the negative heat of the calorimeter because the calorimeter is gaining heat from the reaction. Hence to calculate heat gained by the calorimeter, we use C(delta T). The question asks for the internal energy change for the reaction of 1.00 mol CO. We are given 1.40 g of CO or 0.04998 mol CO. To find out the heat for 1 mol of CO, we divide the value of C(delta T) by 0.04998 mol CO to get -41.2 kJ/mol CO.

If you are trying to find the internal energy change PER mole of CO, why would you divide the value of C(delta T) by 0.04998 mol CO instead of by 1.00 mole of CO?

Diane Bui 2J
Posts: 61
Joined: Sat Jul 22, 2017 3:00 am

### Re: Ch 8 #53 Part b

Hi Timothy,
It's kind of like saying you paid \$20 for 5 apples and trying to find the price of one apple. Since we are given the heat value for 0.04998 mol CO, we divide by that value to get the heat per mol CO.

AKatukota
Posts: 100
Joined: Thu Jul 25, 2019 12:18 am

### Re: Ch 8 #53 Part b

So if the temperature goes up, the reaction is releasing energy and is negative?