# Question 62519

Apr 12, 2015

Here's how you'd go about solving this problem.

Your reaction will initially form liquid water at ${26}^{\circ} \text{C}$. This implies that the released energy will have to account for

• Heating the water from ${26}^{\circ} \text{C}$ to ${100}^{\circ} \text{C}$;
• Convert the water from liquid at ${100}^{\circ} \text{C}$ to vapor at ${100}^{\circ} \text{C}$ - this represents a phase change.
• Heat the water vapor from ${100}^{\circ} \text{C}$ to whatever the final temperature will be.

So, start with the balanced chemical equation for the formation of liquid water

$2 {H}_{2 \left(g\right)} + {O}_{2 \left(g\right)} \to 2 {H}_{2} {O}_{\left(l\right)}$

You know that the standard enthalpy of formation for water is $- \text{285.8 kJ/mol}$. Notice that this value is expressed per mole.

SIDE NOTE The standard enthalpy of formation is defined at 25 degrees Celsius, but since your temperature is very close to that, I'll assume it to be equal to the value measured at standard conditions.

However, your reaction doesn't produce 1 mole, it actually produces 6 moles of water. Because hydrogen gas and water have a $1 : 1$ mole ratio, and since oxygen is not acting as a limiting reagent, the number of moles of water will be equal to the number of moles of hydrogen gas.

So, 6 moles of hydrogen gas react $\to$ 6 moles of water will be produced. As a result, the total heat given off by the reaction will be

DeltaH = 6cancel("moles") * (-"285.8 kJ")/cancel("mol") = -"1714.8 kJ"

Since you have 6 moles of water, you'll have

6cancel("moles") * "18.015 g"/(1cancel("mol")) = "108.1 g of water"

So, to heat water from ${26}^{\circ} \text{C}$ to ${100}^{\circ} \text{C}$, you need

$q = m \cdot c \cdot \Delta T$

q_1 = 108.1cancel("g") * 4.18"J"/(cancel("g") * ^@cancel("C")) * (100 - 26) = "33.44 kJ"

To go from liquid at ${100}^{\circ} \text{C}$ to vapor at ${100}^{\circ} \text{C}$

q_2 = 108.1cancel("g") * 2260"J"/cancel("g") = "244.3 kJ"

From this point on, you'll use the heat given off by the formation reaction to heat the vapor from ${100}^{\circ} \text{C}$ to whatever the final temperature will be.

The remaining amount of enery will be

${q}_{\text{remaining" = q_"given off}} - {q}_{1} - {q}_{2}$

${q}_{\text{remaining" = 1714.8 - 33.44 - 244.3 = "1437.1 kJ}}$

Now solve for the final temperature of the water vapor

q_"remaining" = m * c_"vapor" * (T_"final" - 100)

$1437100 \cancel{\text{J") = 108.1cancel("g") * 2.04cancel("J")/(cancel("g") * ^@cancel("C")) * (T_"final" - 100)^@cancel("C}}$

${T}_{\text{final" = "1459152.4"/220.524 = 6616.8^@"C}}$

Rounded to two sig figs, the number of sig figs given for 6.0 and 3.0 moles, the answer will be

T_"final" = color(green)(6600^@"C")#