Chemistry - Enthelpy Questions (gr 12)

[eleventhxhour]

So I just started learning about enthalpy changes and all the equations that go along with it (for grade 12 chem). I wasn't sure how to do a few questions, so I'd really appreciate the help. I'm mostly kinda confused on when to use the equations q = mcdeltaH and deltaH = ndeltaHx

1) If at 25°C, 15.70g of carbon dioxide absorbs 1.20kJ, determine the final temperature of the gas. Its molar heat capacity is 37.11J/(K mol)

if you have time, I'd appreciate help with this one as well:

2) A 50.0g piece of aluminium is heated to 100.0°C and then put into a beaker containing 150.0mL of water at 20.0°C. Assuming no loss of heat to the surroundings, determine the final temperature of the water.

Thanks!

 

[SuperSonic4]

So I just started learning about enthalpy changes and all the equations that go along with it (for grade 12 chem). I wasn't sure how to do a few questions, so I'd really appreciate the help. I'm mostly kinda confused on when to use the equations q = mcdeltaH and deltaH = ndeltaHx

1) If at 25°C, 15.70g of carbon dioxide absorbs 1.20kJ, determine the final temperature of the gas. Its molar heat capacity is 37.11J/(K mol)

if you have time, I'd appreciate help with this one as well:

2) A 50.0g piece of aluminium is heated to 100.0°C and then put into a beaker containing 150.0mL of water at 20.0°C. Assuming no loss of heat to the surroundings, determine the final temperature of the water.

Thanks!

I suspect you mean $$Q = mc \Delta T$$. That said if you don't know what equation to use I focus on what the question gives me and how the units work together.

1. You can use a similar equation using moles: [math]Q = nc \Delta T[/math]

[math]n = \dfrac{15.7}{44} = 0.357 mol[/math]

Write $$\Delta T = T_f - T_i$$ where $$T_f$$ is the final temperature and [math]T_i[/math] is initial (25)

[math]1200 = 0.357 \cdot 37.11 \cdot (T_f-25)[/math]

Solve for $$T_f$$ which will be in degrees Celsius.

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$$\Delta H = n \Delta H x$$ -- this one doesn't make sense. Should there be a $$T$$ or $$\Delta T$$ in there?

 

[eleventhxhour]

I suspect you mean $$Q = mc \Delta T$$. That said if you don't know what equation to use I focus on what the question gives me and how the units work together.

1. You can use a similar equation using moles: [math]Q = nc \Delta T[/math]

[math]n = \dfrac{15.7}{44} = 0.357 mol[/math]

Write $$\Delta T = T_f - T_i$$ where $$T_f$$ is the final temperature and [math]T_i[/math] is initial (25)

[math]1200 = 0.357 \cdot 37.11 \cdot (T_f-25)[/math]

Solve for $$T_f$$ which will be in degrees Celsius.

---

$$\Delta H = n \Delta H x$$ -- this one doesn't make sense. Should there be a $$T$$ or $$\Delta T$$ in there?

Thanks, I figured out #1! However I'm still having problems with question #2. And the equation I gave you (ΔH=nΔHx) is what my teacher taught us and is using in class (as well as what's in the textbook) The ΔH represents the change of heat. The n is the number of moles. And the ΔHx is the molar enthalpy. So how would you do #2?

 

[SuperSonic4]

Thanks, I figured out #1! However I'm still having problems with question #2. And the equation I gave you (ΔH=nΔHx) is what my teacher taught us and is using in class (as well as what's in the textbook) The ΔH represents the change of heat. The n is the number of moles. And the ΔHx is the molar enthalpy. So how would you do #2?

Ah, I must have missed that. For some reason I imagined the ΔH and x as separate.

2) A 50.0g piece of aluminium is heated to 100.0°C and then put into a beaker containing 150.0mL of water at 20.0°C. Assuming no loss of heat to the surroundings, determine the final temperature of the water.

If there is no loss of heat to the surroundings then the overall heat change must be zero (i.e. an adiabatic process)

[math]m_w c_w (T_f - T_{iw}) + m_a c_a (T_f - T_{ia}) = 0[/math]

where

• $$m_w$$ and $$m_a$$ are the masses of water and aluminium respectively
• $$c_m$$ and $$c_w$$ are the specific heat capacities aluminium [899 J/(K·kg)] and water [4186 J/(K·kg)]
• $$T_f$$ is final temperature
• $$T_{iw}$$ and [math]T_{ia}[/math] are the initial temperatures of the water and aluminium

The density of water does change with temperature but I shall use 1000kg/m^3 for convenience sake

$$m_w = \rho V_w = 1000 \cdot 150 \cdot 10^{-6} = 150 \cdot 10^{-3}\ kg $$

[math]150 \cdot 10^{-3} \cdot 4186 (T_f - 20.0) + 50.0 \cdot 10^{-3} \cdot 889 (T_f - 100) = 0][/math]

Solve for $$T_f$$. As a note I got those figures off google so if your book/data sheet has different values use them

 

[eleventhxhour]

Ah, I must have missed that. For some reason I imagined the ΔH and x as separate.
If there is no loss of heat to the surroundings then the overall heat change must be zero (i.e. an adiabatic process)

[math]m_w c_w (T_f - T_{iw}) + m_a c_a (T_f - T_{ia}) = 0[/math]

where

• $$m_w$$ and $$m_a$$ are the masses of water and aluminium respectively
• $$c_m$$ and $$c_w$$ are the specific heat capacities aluminium [899 J/(K·kg)] and water [4186 J/(K·kg)]
• $$T_f$$ is final temperature
• $$T_{iw}$$ and [math]T_{ia}[/math] are the initial temperatures of the water and aluminium

The density of water does change with temperature but I shall use 1000kg/m^3 for convenience sake

$$m_w = \rho V_w = 1000 \cdot 150 \cdot 10^{-6} = 150 \cdot 10^{-3}\ kg $$

[math]150 \cdot 10^{-3} \cdot 4186 (T_f - 20.0) + 50.0 \cdot 10^{-3} \cdot 889 (T_f - 100) = 0][/math]

Solve for $$T_f$$. As a note I got those figures off google so if your book/data sheet has different values use them

Thanks! I tried that and got the right answer (: