Thread: Solving for heat and deltaH

1. Solving for heat and deltaH

Solving for heat and deltaH?

NaOH + KCl --> NaCl + KOH

Given: 50 ml NaOH in 1.000M,
50ml KCl in 1.000M
Specific heat water: 4.18 J/gC
density of solution: 1.000 g/ml
change in temp: 6.3000C
Find J (heat transferred)
delta H (kJ/mol)

thanks! i have no idea
i only got to the part where you multiply density x 1000 to get mass, but im not even sure thats right.

2. $\displaystyle Q = m \cdot c \cdot \Delta T$

You have a 100 mL solution with a density of 1.000g/mL. This means you have: $\displaystyle (100 \ \text{mL}) \cdot (1.000 \ \text{g mL}^{-1}) = 100 \ \text{g}$ of solution.

So, you have the mass (m), the specific heat capacity of your solution (c, assumed to be that of water), and the change in temperature ($\displaystyle \Delta T$). Plug in your values to get the amount of heat released.

To find $\displaystyle \Delta H$, notice that there are $\displaystyle (50 \ \text{{\color{red}mL}})(0.100 \ \text{mol {\color{blue}L}}^{-1}) = 0.005 \ \text{mol}$ of NaOH or KCl.

This means that for every 0.005 mol of NaOH or KCl used, the value you got for Q is the amount of heat released. So how much heat will be released if exactly 1 mol of NaOH or KCl used? You can then figure out $\displaystyle \Delta H$ (Look at the units, they give you big hints as to what you have to do!)