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High School Chemistry : Help with Enthalpy

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Example Question #1 : Help With Enthalpy

Consider the following balanced reaction:

$2C_{(s)} + 3H_{2}_{(g)} \rightarrow C_{2}H_{6}_{(g)}$

$\Delta H = -84kJ$

What is the change in enthalpy if $\small 10g$ of solid carbon is used in the above reaction?

Possible Answers:

-35kJ

-70kJ

-420kJ

35kJ

Correct answer:

-35kJ

Explanation:

The enthalpy of -84kJ describes the amount of heat when the amount of carbon in the balanced reaction (two moles) is used. Since only $\small 10g$ of carbon are used, we can find how much heat is released.

When two moles of carbon are used, -84kJ are released. Two moles of carbon is equal to $\small 24g$ of carbon, based on carbon's atomic mass.

$2mol\ C*\frac{12g\ C}{1mol\ C}=24g\ C$

Knowing this, we can set up proportions in order to determine how much heat is released by $\small 10g$ of carbon.

$10g\ C*\frac{-84kJ}{24g\ C}=-35kJ$

So $\small 10g$ of carbon results in -35kJ of heat being released to the surroundings.

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Example Question #11 : Thermochemistry And Energetics

Consider these two half reactions:

Step 1. $CH_{4}_{(g)} + 2O_{2}_{(g)} \rightarrow CO_{2}_{(g)} + 2H_{2}O_{(l)}$ $\Delta H = -890kJ$

Step 2. $H_{2}O_{(l)} \rightarrow H_{2}O_{(g)}$ $\Delta H = 44kJ$

Based on these half reactions, find the enthalpy for the following reaction:

$CH_{4}_{(g)} + 2O_{2}_{(g)} \rightarrow CO_{2}_{(g)} + 2H_{2}O_{(g)}$

Possible Answers:

-846kJ

-802kJ

-934kJ

-978kJ

Correct answer:

-802kJ

Explanation:

Hess's law states that the enthalpy of the total reaction is equal to the enthalpy of the steps required to get to the total reaction, regardless of the path that is chosen. This means that we can combine the two half steps with known enthalpies in order to solve for the enthalpy of the main reaction.

Step 1. $CH_{4}_{(g)} + 2O_{2}_{(g)} \rightarrow CO_{2}_{(g)} + 2H_{2}O_{(l)}$ $\Delta H = -890kJ$

Step 2. $H_{2}O_{(l)} \rightarrow H_{2}O_{(g)}$ $\Delta H = 44kJ$

Total: $CH_{4}_{(g)} + 2O_{2}_{(g)} \rightarrow CO_{2}_{(g)} + 2H_{2}O_{(g)}$

Since step 1 results in two moles of liquid water, we need to use the second step twice in order to replace them with two moles of water vapor.

Combined: $CH_{4(g)}+2O_{2(g)}+2H_2O_{(l)} \rightarrow CO_{2(g)}+2H_2O_{(l)}+2H_2O_{(g)}$

Since the total reaction is created by step 1 occurring once and step 2 occurring twice, we can write the enthalpy as:

$\Delta H_{total } = \Delta H_{1} + 2\Delta H_{2}$

Use the given enthalpies of the steps to calculate the total change in enthalpy.

$\Delta H_{total} = -890kJ + 2(44kJ)$

$\Delta H_{total} = -802kJ$

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Example Question #2 : Help With Enthalpy

How much energy is required to heat 35g of H_2O from $35^\circ C$ to $85^\circ C$ ?

Possible Answers:

5.5kJ

7.3kJ

7300kJ

11.2kJ

50J

Correct answer:

7.3kJ

Explanation:

Use the following formula:

$\Delta H =mc \Delta T$

Plug in values:

$\Delta H =35*4.18(85-35)$

$\Delta H =7.3kJ$

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High School Chemistry : Help with Enthalpy

Study concepts, example questions & explanations for High School Chemistry

All High School Chemistry Resources

Example Questions

Example Question #1 : Help With Enthalpy

Example Question #11 : Thermochemistry And Energetics

Example Question #2 : Help With Enthalpy

All High School Chemistry Resources

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