Help with Enthalpy - High School Chemistry

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Question

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?

Answer

The enthalpy of 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, 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 of heat being released to the surroundings.

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Question

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:

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

Answer

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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Question

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

Answer

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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