Question

a chemist measures the energy change (delta h) during the following reaction:

(2no_{2}(g)\to n_{2}o_{4}(g)), (delta h = - 55.3kj)

use the information to answer the following questions.

this reaction is...

suppose (60.8g) of (no_{2}) react.

will any heat be released or absorbed?

if you said heat will be released or absorbed in the second part of this question, calculate how much heat will be released or absorbed.

be sure your answer has the correct number of significant digits.

Explanation:

Step1: Determine the type of reaction

For a reaction, if $\Delta H<0$, it is exothermic (heat is released). Here, $\Delta H = - 55.3\space kJ<0$, so the reaction is exothermic.

Step2: Calculate the moles of $NO_2$

The molar mass of $NO_2$ is $M=(14 + 2\times16)\space g/mol=46\space g/mol$.
The number of moles of $NO_2$, $n=\frac{m}{M}$, where $m = 60.8\space g$. So $n=\frac{60.8\space g}{46\space g/mol}\approx1.322\space mol$.

Step3: Relate moles of $NO_2$ to $\Delta H$

From the reaction $2NO_2(g)\to N_2O_4(g)\space \Delta H=- 55.3\space kJ$, for $2\space mol$ of $NO_2$ reacting, the heat change is $-55.3\space kJ$.
Let $x$ be the heat change for $n = 1.322\space mol$ of $NO_2$ reacting. Using the proportion $\frac{x}{1.322}=\frac{-55.3}{2}$.
Then $x=\frac{-55.3\times1.322}{2}\space kJ\approx - 36.6\space kJ$. The negative sign indicates heat is released.

Answer:

• This reaction is: exothermic.
• Suppose $60.8\space g$ of $NO_2$ react. Will any heat be released or absorbed?: Yes, released.
• If you said heat will be released or absorbed in the second part of this question, calculate how much heat will be released or absorbed: $-36.6\space kJ$