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Experiment A
3cm metal
0.5M acid
5 degrees temperature change
Experiment B
6cm metal
1M acid
How much would the temperature change be, and how much faster or slower would the reaction take place in?
I thought the answer would be either "the same" or "4x as much". I don't exactly understand the reaction yet; would the doubling of each material make the reaction 4x stronger (2x2), or would they cancel each other out?
Of course, it might've been something else I haven't thought of, too.
Without providing more detailed specifications of the metal such as: state of the metal - powder or solid, shape of the solid form etc., the rate of the reaction will be the same, but the second reaction will be a bit faster because of the bigger temperature change (as it affects reaction rate coefficient).
I assume that you meant 3 and 6 $cm^3$ and not $cm$. The same metal reacts in both reactions. So we can assume that the density of both metals is also the same ($\rho_1=\rho_2=\rho$). We can further assume that metals were dissolved completely and that the temperature change is actually temperature increase.
$\rho=\frac{m_1}{V_1}$, $\rho=\frac{m_2}{V_2}$
$\frac{m_1}{V_1}=\frac{m_2}{V_2}$
$\frac{m_1}{3}=\frac{m_2}{6}$
$\frac{m_1}{1}=\frac{m_2}{2}$
The bigger piece of metal weights 2 times more than the smaller metal. We can convert weights to moles (molar mass of both metals is the same):
$n=\frac{m}{M}$
$\frac{n_1}{1}=\frac{n_2}{2}$ so we can say $2*n_1=n_2$
The 1st reaction:
$n_1$ reacts with 0,5M of acid.
The 2nd reaction:
2*$n_1$ react with 1M solution and it is the same as $n_1$ reacts with 0,5M
The volume of the solution is the same and the specific heat capacity will be nearly the same. The reaction enthalpy corresponds to the amount of reacted material. Because two times more material reacts in the second reaction, the temperature change will be nearly two times bigger (nearly - because of the change in specific heat capacity).
