Chemical EquilibriumHard

Question

For the reaction: CuSO4.3H2O(s) $\rightleftharpoons$CuSO4.H2O(s) + 2H2O(g); ΔH = 3360 cal. The dissociation pressure is 7×10–3 atm at 27oC. What will be the dissociation pressure at 127oC (ln2 = 0.7)?

Options

A.9.8 × 10–3 atm
B.1.4 × 10–2 atm
C.1.4 × 10–3 atm
D.9.8 × 10–2 atm

Solution

$K_{p} = P^{2}$

Now, $\ln\frac{K_{2}}{K_{1}} = \frac{\Delta H}{R}\left( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right)$

Or, $\ln\frac{P_{2}^{2}}{\left( 7 \times 10^{- 3} \right)^{2}} = \frac{3360}{2}\left( \frac{1}{300} - \frac{1}{400} \right)$

$\Rightarrow P_{2} = 1.4 \times 10^{- 2}\text{ atm}$

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