Question

A coffee-cup (constant pressure) calorimeter is used to carry out the following reaction in an unknown volume of water (where X is a hypothetical metal):

X + 2 H2O → X(OH)2 + H2

In this process, the water temperature rose from 25.0 °C to 32.2 °C. If 0.00803 mol of "X" was consumed during the reaction, and the ΔH of this reaction with respect to the system is -1798 kJ mol-1 , what volume of water (in mL) was present in the calorimeter?

The specific heat of water is 4.184 J g-1 °C-1

Answer 1

Task #82213

A coffee-cup (constant pressure) calorimeter is used to carry out the following reaction in an unknown volume of water (where X is a hypothetical metal):

X + 2 H2O → X(OH)2 + H2

In this process, the water temperature rose from $25.0^{\circ}\mathrm{C}$ to $32.2^{\circ}\mathrm{C}$. If 0.00803 mol of "X" was consumed during the reaction, and the $\Delta H$ of this reaction with respect to the system is -1798 kJ mol-1, what volume of water (in mL) was present in the calorimeter?

The specific heat of water is 4.184 \, \text{J g}^{-1} \, ^{\circ}\text{C}^{-1}.

Solution.

Firstly, we should find heat that was released by the reaction, using $\Delta H$.

$$1 \, \text{mole} \, (^{\prime \prime} X^{\prime \prime}) - 1798 \, \text{kJ}$$ $$0.00803 \, \text{mole} \, (^{\prime \prime} X^{\prime \prime}) - x \, \text{kJ}$$ $$x = 0.00803 \, \text{mole} \times 1798 \, \text{kJ}/1 \, \text{mole} = 14.44 \, \text{kJ}$$ $$x - Q, \, \text{which was released}.$$ $$Q = Q_{\text{water}}$$ $$Q = c \cdot m \cdot \Delta t, \, \Delta t = 32.2^{\circ} \text{C} - 25.0^{\circ} \text{C} = 7.2^{\circ} \text{C}$$ $$Q = c \cdot \rho \cdot V \cdot \Delta t$$ $$V = Q / c \cdot \rho \cdot \Delta t$$ $$V = 14.44 \, \text{kJ} / 4.184 \, (\text{J/g} \cdot ^{\circ} \text{C}) \cdot 1 \, \text{g/cm}^3 \cdot 7.2^{\circ} \text{C} = 479.44 \, \text{cm}^3 = 479.44 \, \text{ml}$$

Answer:

$$V = 14.44 \, \text{kJ} / 4.184 \, (\text{J/g} \cdot ^{\circ} \text{C}) \cdot 1 \, \text{g/cm}^3 \cdot 7.2^{\circ} \text{C} = 479.44 \, \text{cm}^3 = 479.44 \, \text{ml}$$

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