Chemistry, 28.05.2020 22:02 shay7298

Question 1

If we mixed 45cm^3 of 2m NaOH with 20cm^3 of 1m HCL and obtained a temperature difference of 9.4°C, what would be the heat of reaction, the enthalpy?.

Question 2

If we added 10cm^3 HCL of an unknown concentration to 30cm^3 of 2M NaOH and obtained a temperature change of 4.5°C and then added 20cm^3 of the same unknown acid to 30cm^3 of 2M NaOH and obtained a temperature change of 9.0°C, what is the concentration of the acid? Use the enthalpy value from question 1

Answers: 1

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Answer from: hamaama

The heat of reaction is $Q = 170.92 \ J$

The enthalpy is $\Delta H = -170.91 \ J$

The concentration of HCl is $M__{HCl}} =2.468 M$

Explanation:

From the question we are told that

The volume of NaOH is $V_{NaOH} = 45 \ cm^3 = \frac{45} {1000} = 0.045 L$

The number of concentration of NaOH is $M__{NaOH}} = 2M$

The volume of HCl is $V_{HCl} = 20 cm^3$

The number of concentration of HCl is $M__{HCl}} = 1 M$

The temperature difference is $\Delta T = 9.4 ^o C$

Now the heat of reaction is mathematically represented as

$Q = m * c_p * \Delta T$

Where

c_p is the specific heat of water with value c_p = 4200 J /kg ^o C

$m = m__{NaOH}} + m__{HCl}}$

Now $m__{NaOH}} = V_{NaOH} * M_{NaOH} * Z_{NaOH}$

where $Z_{NaOH}$ is the molar mass of NaOH with the value of 0.04 kg/mol

So $m__{NaOH}} = 0.045 * 2 * 0.040$

$m__{NaOH}} =0.0036\ kg$

While

$m__{HCl}} = V_{HCl} * M_{HCl} + Z_{HCl}$

Where $Z_{HCl}$ is the molar mass of HCl with the value of 0.03646 kg/mol

$m__{NaOH}} = 0.020 * 1 * 0.03646$

$m__{NaOH}} = 0.000729 kg$

m = 0.0036 + 0.000729

m = 0.00433

=> Q = 0.00433 * 4200 * 9.4

$Q = 170.92 \ J$

The enthalpy is mathematically represented as

$\Delta H = - Q$

=> $\Delta H = -170.91 \ J$

From the second question we are told that

The volume of HCl is $V_{HCl_1} = 10cm^3 = \frac{10}{1000} = 0.010 L$

The volume of NaOH is $V_{NaOH_1 } = 30 cm^3 = \frac{30}{1000} = 0.03 L$

The concentration of NaOH is $M_{NaOH} = 2 M$

The first temperature change is $\Delta T = 4.5 ^oC$

The second volume of $V_{HCl_2} = 20 cm^3 = \frac{20}{1000 } = 0.020 m^3$

The mass of NaOH is

$m__{NaOH}} = V_{NaOH} * M_{NaOH} * Z_{NaOH}$

substituting values

$m__{NaOH}} = 0.03 * 2 * 40$

$m__{NaOH}} = 3.6 \ g$

The mass of the product formed is

$m = \frac{Q}{c_p * \Delta T}$

substituting values

$m = \frac{170.91}{4200 * 9} * 1000$

The multiplication by 1000 is to convert it from kg to grams

m = 4.5 g

Now the mass of HCl is

$m__{HCl}} = m - m__{NaOh}}$

substituting values

$m__{HCl}} = 4.5 -3.6$

$m__{HCl}} = 0.9 \ g$

Now the concentration of HCl is

$M__{HCl}} = \frac{m_{HCl}}{(Z_{HCl} * *1000) * V_{HCl_1}}}$

The multiplication of $Z_{HCl}$ is to convert it from kg/mol to g/mol

$M__{HCl}} = \frac{0.9}{36.46 * 0.01}}$

$M__{HCl}} =2.468 M$

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